RS Aggarwal Solutions for Class 11 Chapter 09 Combinations

Question 1. Evaluate: \( ^{20}C_4 \)
Answer: We understand that the combination formula is \( ^nC_r = \frac{n!}{(n-r)! \times r!} \).

\( ^{20}C_4 = \frac{20!}{(20-4)! \times 4!} \)

\( ^{20}C_4 = \frac{20!}{16! \times 4!} \)

\( ^{20}C_4 = \frac{20 \times 19 \times 18 \times 17 \times 16!}{16! \times 4!} \)

\( ^{20}C_4 = \frac{20 \times 19 \times 18 \times 17}{4!} \)

\( ^{20}C_4 = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} \)

\( ^{20}C_4 = \frac{116280}{24} \)

\( ^{20}C_4 = 4845 \)

Exam Tip: Always cancel common factors before multiplying large numbers - this helps reduce arithmetic errors and makes calculations faster.

Question 2. Evaluate: \( ^{16}C_{13} \)
Answer: We understand that \( ^nC_r = \frac{n!}{(n-r)! \times r!} \).

\( ^{16}C_{13} = \frac{16!}{(16-13)! \times 13!} \)

\( ^{16}C_{13} = \frac{16!}{3! \times 13!} \)

\( ^{16}C_{13} = \frac{16 \times 15 \times 14 \times 13!}{3! \times 13!} \)

\( ^{16}C_{13} = \frac{16 \times 15 \times 14}{3!} \)

\( ^{16}C_{13} = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} \)

\( ^{16}C_{13} = \frac{3360}{6} \)

\( ^{16}C_{13} = 560 \)

Exam Tip: Use the symmetry property \( ^nC_r = ^nC_{n-r} \) to simplify - in this case, \( ^{16}C_{13} = ^{16}C_3 \) makes the calculation much shorter.

Question 3. Evaluate: \( ^{90}C_{88} \)
Answer: We understand that \( ^nC_r = \frac{n!}{(n-r)! \times r!} \).

\( ^{90}C_{88} = \frac{90!}{(90-88)! \times 88!} \)

\( ^{90}C_{88} = \frac{90!}{2! \times 88!} \)

\( ^{90}C_{88} = \frac{90 \times 89 \times 88!}{2! \times 88!} \)

\( ^{90}C_{88} = \frac{90 \times 89}{2!} \)

\( ^{90}C_{88} = \frac{90 \times 89}{2 \times 1} \)

\( ^{90}C_{88} = \frac{8010}{2} \)

\( ^{90}C_{88} = 4005 \)

Exam Tip: When r is large relative to n, apply the symmetry property \( ^nC_r = ^nC_{n-r} \) immediately to work with smaller numbers and reduce effort.

Question 4. Evaluate: \( ^{71}C_{71} \)
Answer: We understand that \( ^nC_r = \frac{n!}{(n-r)! \times r!} \).

\( ^{71}C_{71} = \frac{71!}{(71-71)! \times 71!} \)

\( ^{71}C_{71} = \frac{71!}{0! \times 71!} \)

\( ^{71}C_{71} = \frac{1}{0!} \) (since 0! = 1)

\( ^{71}C_{71} = 1 \)

Exam Tip: Remember that \( ^nC_n = 1 \) always - there is exactly one way to choose all n items from n items.

Question 5. Evaluate: \( ^{n+1}C_n \)
Answer: We understand that \( ^nC_r = \frac{n!}{(n-r)! \times r!} \).

\( ^{n+1}C_n = \frac{(n+1)!}{(n+1-n)! \times n!} \)

\( ^{n+1}C_n = \frac{(n+1)!}{1! \times n!} \)

\( ^{n+1}C_n = \frac{(n+1) \times n!}{1 \times n!} \) (since 1! = 1)

\( ^{n+1}C_n = \frac{(n+1)}{1} \)

\( ^{n+1}C_n = n+1 \)

Exam Tip: Recognize the pattern: \( ^nC_{n-1} = n \) and \( ^{n+1}C_n = n+1 \) - these appear frequently in combination problems and save calculation time.

Question 6. Evaluate: \( \sum_{r=1}^{6} \binom{6}{r} \)
Answer: We understand that \( \sum_{r=1}^{n} \binom{n}{r} = 2^n - \binom{n}{0} \).

\( \sum_{r=1}^{6} \binom{6}{r} = 2^6 - \binom{6}{0} \)

\( \sum_{r=1}^{6} \binom{6}{r} = 64 - 1 \)

\( \sum_{r=1}^{6} \binom{6}{r} = 63 \)

Exam Tip: Recall that the sum of all binomial coefficients from 0 to n equals \( 2^n \) - when you exclude the first term, subtract 1 from this total.

Question 7. Verify that: (i) \( ^{15}C_8 + ^{15}C_9 - ^{15}C_6 - ^{15}C_7 = 0 \)
Answer: Given: \( ^{15}C_8 + ^{15}C_9 - ^{15}C_6 - ^{15}C_7 \)

To prove: \( ^{15}C_8 + ^{15}C_9 - ^{15}C_6 - ^{15}C_7 = 0 \)

We apply the symmetry property: \( ^nC_r = ^nC_{n-r} \)

\( ^{15}C_8 = ^{15}C_7 \) and \( ^{15}C_9 = ^{15}C_6 \)

Substituting these relationships:
\( ^{15}C_8 + ^{15}C_9 - ^{15}C_6 - ^{15}C_7 = ^{15}C_8 + ^{15}C_9 - ^{15}C_9 - ^{15}C_8 = 0 \)

Hence verified that \( ^{15}C_8 + ^{15}C_9 - ^{15}C_6 - ^{15}C_7 = 0 \)

Exam Tip: Always check whether symmetry can simplify the expression - pairing equal combinations often leads to cancellation and fast verification.

Question 8. Verify that: (ii) \( ^{10}C_4 + ^{10}C_3 = ^{11}C_4 \)
Answer: We apply the combination addition rule: \( ^nC_r + ^nC_{r-1} = ^{n+1}C_r \)

Here, n = 10 and r = 4.

L.H.S. = \( ^{10}C_4 + ^{10}C_3 = ^{11}C_4 \)

Hence verified.

Exam Tip: The Pascal's triangle rule \( ^nC_r + ^nC_{r-1} = ^{n+1}C_r \) is fundamental - memorize it as it appears in most combination proofs.

Question 9. (i) If \( ^nC_7 = ^nC_5 \), find n.
Answer: Given: \( ^nC_7 = ^nC_5 \)

To find: n

We apply the property: \( ^nC_r = ^nC_{n-r} \)

\( ^nC_7 = ^nC_{n-7} \)

Since \( ^nC_{n-7} = ^nC_5 \)

\( n - 7 = 5 \)

\( n = 12 \)

Exam Tip: When \( ^nC_r = ^nC_s \), either r = s or r + s = n - use both possibilities to avoid missing solutions.

Question 10. (ii) If \( ^nC_{14} = ^nC_{16} \), find \( ^nC_{28} \).
Answer: Given: \( ^nC_{14} = ^nC_{16} \)

To find: \( ^nC_{28} \)

We apply the property: \( ^nC_r = ^nC_{n-r} \)

\( ^nC_{14} = ^nC_{n-14} \)

Since \( ^nC_{n-14} = ^nC_{16} \)

\( n - 14 = 16 \)

\( n = 30 \)

Now, \( ^nC_{28} = ^{30}C_{28} = \frac{30!}{(30-28)! \times 28!} = \frac{30!}{2! \times 28!} = \frac{30 \times 29}{2 \times 1} = \frac{870}{2} = 435 \)

Exam Tip: First determine n from the equality condition, then compute the requested combination value using the formula.

Question 11. (iii) If \( ^nC_{16} = ^nC_{14} \), find \( ^nC_{27} \).
Answer: Given: \( ^nC_{16} = ^nC_{14} \)

To find: \( ^nC_{27} \)

We apply the property: \( ^nC_r = ^nC_{n-r} \)

\( ^nC_{14} = ^nC_{n-14} \)

Since \( ^nC_{n-14} = ^nC_{16} \)

\( n - 14 = 16 \)

\( n = 30 \)

Now, \( ^nC_{27} = ^{30}C_{27} = \frac{30!}{(30-27)! \times 27!} = \frac{30!}{3! \times 27!} = \frac{30 \times 29 \times 28 \times 27!}{6 \times 27!} = \frac{30 \times 29 \times 28}{6} = \frac{24360}{6} = 4060 \)

Exam Tip: After finding n, verify that the requested combination value is valid - ensure that the r value does not exceed n.

Question 12. (i) If \( ^{20}C_r = ^{20}C_{r+6} \), find r.
Answer: Given: \( ^{20}C_r = ^{20}C_{r+6} \)

To find: r

We apply the property: \( ^nC_r = ^nC_{n-r} \)

\( ^{20}C_{r+6} = ^{20}C_{20-(r+6)} = ^{20}C_{14-r} \)

Since \( ^{20}C_{14-r} = ^{20}C_r \)

\( 14 - r = r \)

\( 2r = 14 \)

\( r = 7 \)

Exam Tip: When two combinations with the same n are equal, set either the subscripts equal or use the symmetry property to create an equation.

Question 13. (ii) If \( ^{18}C_r = ^{18}C_{r+2} \), find \( ^rC_5 \).
Answer: Given: \( ^{18}C_r = ^{18}C_{r+2} \)

To find: \( ^rC_5 \)

We apply the property: \( ^nC_r = ^nC_{n-r} \)

\( ^{18}C_{r+2} = ^{18}C_{18-(r+2)} = ^{18}C_{16-r} \)

Since \( ^{18}C_{16-r} = ^{18}C_r \)

\( 16 - r = r \)

\( 2r = 16 \)

\( r = 8 \)

Now, \( ^rC_5 = ^8C_5 = \frac{8!}{(8-5)! \times 5!} = \frac{8!}{3! \times 5!} = \frac{8 \times 7 \times 6 \times 5!}{6 \times 5!} = \frac{8 \times 7 \times 6}{6} = 8 \times 7 = 56 \)

Exam Tip: Break multi-step problems into parts - first find the variable, then use that result to calculate the final answer.

Question 14. If \( ^nC_{r-1} = ^nC_{3r} \), find r.
Answer: Given: \( ^nC_{r-1} = ^nC_{3r} \)

To find: r

We apply the property: \( ^nC_r = ^nC_{n-r} \)

\( ^nC_{r-1} = ^nC_{n-(r-1)} = ^nC_{n-r+1} \)

Since \( ^nC_{n-r+1} = ^nC_{3r} \)

\( n - r + 1 = 3r \)

\( 4r = n + 1 \)

\( r = \frac{n+1}{4} \)

Exam Tip: Express the answer in simplest form - if the answer contains n, leave it as an algebraic expression unless additional constraints are given.

Question 15. If \( ^{2n}C_3 : ^nC_3 = 12 : 1 \), find n.
Answer: Given: \( ^{2n}C_3 : ^nC_3 = 12 : 1 \)

To find: n

\( \frac{^{2n}C_3}{^nC_3} = \frac{12}{1} \)

\( \frac{\frac{(2n)!}{(2n-3)! \times 3!}}{\frac{n!}{(n-3)! \times 3!}} = 12 \)

\( \frac{(2n)! \times (n-3)!}{(2n-3)! \times n!} = 12 \)

\( \frac{(2n) \times (2n-1) \times (2n-2) \times (2n-3)! \times (n-3)!}{(2n-3)! \times n \times (n-1) \times (n-2) \times (n-3)!} = 12 \)

\( \frac{(2n) \times (2n-1) \times (2n-2)}{n \times (n-1) \times (n-2)} = 12 \)

\( \frac{2n \times (2n-1) \times (n-1)}{n \times (n-1) \times (n-2)} = 12 \)

\( \frac{2(2n-1)}{(n-2)} = 12 \)

\( 2(2n-1) = 12(n-2) \)

\( 4n - 2 = 12n - 24 \)

\( 22 = 8n \)

\( 2n - 1 = 3(n - 2) \)

\( 2n - 1 = 3n - 6 \)

\( n = 5 \)

Exam Tip: When working with ratios of combinations, expand both numerator and denominator fully, then simplify before applying the given ratio.

Question 16. If \( ^{15}C_r : ^{15}C_{r-1} = 11 : 5 \), find r.
Answer: Given: \( ^{15}C_r : ^{15}C_{r-1} = 11 : 5 \)

To find: r

\( \frac{^{15}C_r}{^{15}C_{r-1}} = \frac{11}{5} \)

\( \frac{\frac{15!}{(15-r)! \times r!}}{\frac{15!}{(15-r+1)! \times (r-1)!}} = \frac{11}{5} \)

\( \frac{15! \times (15-r+1)! \times (r-1)!}{(15-r)! \times r! \times 15!} = \frac{11}{5} \)

\( \frac{(16-r) \times (r-1)!}{r \times (r-1)! \times (15-r)!} \times (15-r)! = \frac{11}{5} \)

\( \frac{(16-r)}{r} = \frac{11}{5} \)

\( 5(16-r) = 11r \)

\( 80 - 5r = 11r \)

\( 80 = 16r \)

\( r = 5 \)

Exam Tip: Simplify the ratio of consecutive combinations directly using the relationship \( \frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r} \) to save time.

Question 17. If \( ^nP_r = 840 \) and \( ^nC_r = 35 \), find the value of r.
Answer: Given: \( ^nP_r = 840 \) and \( ^nC_r = 35 \)

To find: r

We use the relationship: \( ^nP_r = ^nC_r \times r! \)

\( 840 = 35 \times r! \)

\( r! = \frac{840}{35} = 24 \)

\( r! = 4! \)

\( r = 4 \)

Exam Tip: Always use the formula \( ^nP_r = ^nC_r \times r! \) to connect permutations and combinations - this often simplifies the problem significantly.

Question 18. If \( ^nC_{r-1} = 36 \), \( ^nC_r = 84 \) and \( ^nC_{r+1} = 126 \), find r.
Answer: Given: \( ^nC_{r-1} = 36 \), \( ^nC_r = 84 \) and \( ^nC_{r+1} = 126 \)

To find: r

Using the ratio: \( \frac{^nC_r}{^nC_{r-1}} = \frac{84}{36} = \frac{7}{3} \)

\( \frac{n!/(n-r)! \times r!}{n!/(n-r+1)! \times (r-1)!} = \frac{7}{3} \)

\( \frac{(n-r+1)}{r} = \frac{7}{3} \)

\( 3(n-r+1) = 7r \)

\( 3n - 10r = -3 \) ... (1)

Using the ratio: \( \frac{^nC_{r+1}}{^nC_r} = \frac{126}{84} = \frac{3}{2} \)

\( \frac{(n-r)}{(r+1)} = \frac{3}{2} \)

\( 2(n-r) = 3(r+1) \)

\( 2n - 5r = 3 \) ... (2)

Solving equations (1) and (2):
From equation (1): \( 3n = 10r - 3 \)
Substituting in equation (2): \( 2 \times \frac{10r-3}{3} - 5r = 3 \)
\( \frac{20r - 6}{3} - 5r = 3 \)
\( 20r - 6 - 15r = 9 \)
\( 5r = 15 \)
\( r = 3 \), and \( n = 9 \)

Exam Tip: When given three consecutive combination values, create two ratio equations and solve the system - this eliminates n efficiently.

Question 19. If \( ^{n+1}C_{r+1} : ^nC_r = 11 : 6 \) and \( ^nC_r : ^{n-1}C_{r-1} = 6 : 3 \), find n and r.
Answer: Given: \( ^{n+1}C_{r+1} : ^nC_r = 11 : 6 \) and \( ^nC_r : ^{n-1}C_{r-1} = 6 : 3 \)

To find: n and r

From the first ratio:
\( \frac{^{n+1}C_{r+1}}{^nC_r} = \frac{11}{6} \)

Using the property \( ^nC_r = \frac{n}{r} \times ^{n-1}C_{r-1} \):

\( \frac{(n+1)!/(n+1-r-1)! \times (r+1)!}{n!/(n-r)! \times r!} = \frac{11}{6} \)

\( \frac{(n+1)}{(r+1)} = \frac{11}{6} \)

\( 6(n+1) = 11(r+1) \)

\( 6n + 6 = 11r + 11 \)

\( 6n - 11r = 5 \) ... (1)

From the second ratio:
\( \frac{^nC_r}{^{n-1}C_{r-1}} = \frac{6}{3} = 2 \)

\( \frac{n!/(n-r)! \times r!}{(n-1)!/(n-1-r+1)! \times (r-1)!} = 2 \)

\( \frac{n}{r} = 2 \)

\( n = 2r \) ... (2)

Substituting (2) into (1):
\( 6(2r) - 11r = 5 \)
\( 12r - 11r = 5 \)
\( r = 5 \)

\( n = 2 \times 5 = 10 \)

Exam Tip: Recognize the factorization patterns in combination ratios - the property \( \frac{^nC_r}{^{n-1}C_{r-1}} = \frac{n}{r} \) simplifies many ratio problems quickly.

Question 20. How many different teams of 11 players can be chosen from 15 players?
Answer: Condition: Each player has an equal opportunity to be selected.

Picture picking team members sequentially. The first selection offers 15 options, the second presents 14 options, the third gives 13 options, and continuing this pattern down to 5 options for the eleventh selection.

This problem involves finding combinations since the arrangement order is not relevant.

\( n = 15 \) and \( r = 11 \)

\( ^nC_r = ^{15}C_{11} \)

\( ^{15}C_{11} = \frac{15!}{(15-11)! \times 11!} \)

\( ^{15}C_{11} = \frac{15!}{4! \times 11!} \)

\( ^{15}C_{11} = \frac{15 \times 14 \times 13 \times 12 \times 11!}{4! \times 11!} \)

\( ^{15}C_{11} = \frac{15 \times 14 \times 13 \times 12}{24} = \frac{32760}{24} = 1365 \)

Therefore, 1365 different teams can be formed.

Exam Tip: Always recognize when order does not matter (use combinations) versus when order matters (use permutations) - the word "team" typically indicates combinations since team membership order is irrelevant.

Question 12. If there are 12 persons in a party and if each two of them shake hands with each other, how many handshakes are possible?
Answer: When 12 people come together, we need to pick a group of two different people where the order in which we select them doesn't matter. Since each person shakes hands with every other person exactly once, we need to find all such pairs. The formula we use is: nCr where n=12 and r=2.

nCr= 12C2

\( ^{12}C_2 = \frac{12!}{(12-2)! \times 2!} \)

\( ^{12}C_2 = \frac{12!}{10! \times 2!} \)

\( ^{12}C_2 = \frac{12 \times 11 \times 10!}{10! \times 2!} \)

\( ^{12}C_2 = \frac{12 \times 11}{2 \times 1} \)

\( ^{12}C_2 = 66 \)

In total, 66 handshakes can happen at a party of 12 people when each pair exchanges a handshake exactly once.
In simple words: Pick any two people from 12. The number of different pairs you can make is 66. Each pair shakes hands once, so there are 66 handshakes.

Exam Tip: Always identify whether order matters (permutation) or not (combination). Here, shaking hands with person A then person B is the same as shaking hands with B then A - so use combination (nCr), not permutation.

Question 13. How many chords can be drawn through 21 points on a circle?
Answer: We have 21 points on the circle.

⇒ n=21

A chord is formed by connecting any two points on the circle.

⇒ r=2

The total number of chords from 21 points = nCr

⇒ nCr = 21C2

\( ^{21}C_2 = \frac{21!}{(21-2)! \times 2!} \)

\( ^{21}C_2 = \frac{21!}{19! \times 2!} \)

\( ^{21}C_2 = \frac{21 \times 20 \times 19!}{19! \times 2!} \)

\( ^{21}C_2 = \frac{21 \times 20}{2 \times 1} \)

\( ^{21}C_2 = 210 \)

210 chords can be drawn through 21 points on a circle.
In simple words: To draw a chord, you need to pick any 2 points on the circle. From 21 points, the number of ways to pick 2 points is 210, so there are 210 possible chords.

Exam Tip: Remember that a chord always requires exactly two endpoints. The question is really asking: "In how many ways can you choose 2 points from 21?" That's a combination problem.

Question 14. From a class of 25 students, 4 are to be chosen for a competition. In how many ways can this be done?
Answer: This is a combination problem where we select items without regard to order.

Here, n=25 and r=4

⇒ nCr = 25C4

\( ^{25}C_4 = \frac{25!}{(25-4)! \times 4!} \)

\( ^{25}C_4 = \frac{25!}{21! \times 4!} \)

\( ^{25}C_4 = \frac{25 \times 24 \times 23 \times 22 \times 21!}{21! \times 4!} \)

\( ^{25}C_4 = \frac{25 \times 24 \times 23 \times 22}{4 \times 3 \times 2 \times 1} \)

\( ^{25}C_4 = 12650 \)

In 12,650 ways, 4 students can be picked from a class of 25 for the competition.
In simple words: You need to choose 4 students from 25. The order in which you pick them doesn't matter (student A picked first or second is the same group). The total number of different groups you can make is 12,650.

Exam Tip: Since the question asks "In how many ways can this be done?" with no mention of positions or roles, this is a combination problem. If the 4 students were assigned different roles (captain, vice-captain, etc.), it would be a permutation problem instead.

Exercise 9B

Question 1. In how many ways can 5 sportsmen be selected from a group of 10?
Answer: We have 10 sportsmen from which 5 need to be selected.

5 sportsmen can be picked from 10 using the combination formula 10C5.

Applying nCr = \( \frac{n!}{r!(n-r)!} \)

We get:

\( ^{10}C_5 = \frac{10!}{5!(10-5)!} = \frac{10!}{5! \times 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \)

Therefore, 252 ways exist for selecting 5 sportsmen from a group of 10 sportsmen.
In simple words: Pick 5 people from 10. The number of different groups of 5 you can make is 252.

Exam Tip: Always check if the problem involves selecting (combination) or arranging (permutation). "Selected" typically means combination - the order doesn't matter.

Question 2. A bag contains 5 black and 6 red balls. Find the number of ways in which 2 black and 3 red balls can be selected.
Answer: The bag holds 5 black balls and 6 red balls. We need to find how many ways we can pick 2 black and 3 red balls.

The number of ways to choose 2 black balls from 5 black balls is 5C2, and the number of ways to choose 3 red balls from 6 red balls is 6C3.

Using the multiplication principle (since we are doing both selections), the total number of ways will be 5C2 × 6C3.

Applying nCr = \( \frac{n!}{r!(n-r)!} \):

\( ^5C_2 = \frac{5!}{2! \times 3!} = \frac{5 \times 4}{2 \times 1} = 10 \)

\( ^6C_3 = \frac{6!}{3! \times 3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \)

Total = 10 × 20 = 200

The total number of ways to select 2 black and 3 red balls is 200.
In simple words: First, pick 2 black from 5 (10 ways). Then pick 3 red from 6 (20 ways). Since you do both, multiply: 10 × 20 = 200 total ways.

Exam Tip: When a problem says "select A from group 1 AND B from group 2," use the multiplication principle. Multiply the number of ways for each separate selection.

Question 3. Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 4 blue balls if each selection consists of 3 balls of each colour.
Answer: We have a total of 6 red balls, 5 white balls, and 4 blue balls. We must pick 3 balls from each colour group.

Number of ways to pick 3 red balls from 6 red balls = 6C3

Number of ways to pick 3 white balls from 5 white balls = 5C3

Number of ways to pick 3 blue balls from 4 blue balls = 4C3

Using the multiplication principle, the total number of ways would be:

⇒ 6C3 × 5C3 × 4C3

Applying formula, nCr = \( \frac{n!}{r!(n-r)!} \), we get

\( ^6C_3 = \frac{6!}{3! \times 3!} = 20 \)

\( ^5C_3 = \frac{5!}{3! \times 2!} = 10 \)

\( ^4C_3 = \frac{4!}{3! \times 1!} = 4 \)

Total = 20 × 10 × 4 = 800

The number of ways to select 9 balls from 6 red, 5 white and 4 blue balls with exactly 3 of each colour is 800.
In simple words: Choose 3 from red (20 ways), 3 from white (10 ways), and 3 from blue (4 ways). Multiply all three: 20 × 10 × 4 = 800.

Exam Tip: When selecting from multiple groups with specific requirements for each group, calculate combinations for each group separately, then multiply using the multiplication principle.

Question 4. How many different boat parties of 8 consisting of 5 boys and 3 girls can be made from 20 boys and 10 girls?
Answer: We need to form a boat party of 8 people - 5 must be boys and 3 must be girls. We have 20 boys and 10 girls to choose from.

Number of ways to choose 5 boys from 20 boys = 20C5

\( ^{20}C_5 = \frac{20!}{5! \times 15!} = \frac{20 \times 19 \times 18 \times 17 \times 16}{5 \times 4 \times 3 \times 2 \times 1} = 15,504 \)

Number of ways to choose 3 girls from 10 girls = 10C3

\( ^{10}C_3 = \frac{10!}{3! \times 7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \)

Using the multiplication principle, the total number of ways = 20C5 × 10C3 = 15,504 × 120 = 1,860,480
In simple words: Pick 5 boys from 20 (15,504 ways) and 3 girls from 10 (120 ways). Multiply to get the total number of different boat parties: 1,860,480.

Exam Tip: Break the problem into independent selections. Select boys and girls separately, then use the multiplication principle to find the total - don't try to combine them first.

Question 5. In how many ways can a student choose 5 courses out of 9 courses if 2 specific courses are compulsory for every student?
Answer: Since every student must choose 5 courses total and 2 are already mandatory, the student only needs to pick 3 more courses from the remaining 7 courses.

Number of ways to choose 3 additional courses from 7 available courses = 7C3

Applying formula, nCr = \( \frac{n!}{r!(n-r)!} \):

\( ^7C_3 = \frac{7!}{3! \times 4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \)

The student can choose 5 courses in 35 different ways.
In simple words: 2 courses are forced choices. You pick the remaining 3 from 7 optional courses. That can be done in 35 ways.

Exam Tip: When certain items are compulsory, reduce both the total number of items to choose and the number you need to select. Here: choose 3 from 7 (not 5 from 9).

Question 6. A sports team of 11 students is to be constituted, choosing at least 5 from class XI and at least 5 from class XII. If there are 20 students in each of these classes, in how many ways can the team be constituted?
Answer: We need to form a team of 11 with at least 5 from each class, and there are 20 students in each class.

There are two possible ways the selection can happen:

**Case 1:** Picking 5 from class XI and 6 from class XII

Number of ways = 20C5 × 20C6

**Case 2:** Picking 6 from class XI and 5 from class XII

Number of ways = 20C6 × 20C5

Total number of ways = (20C5 × 20C6) + (20C6 × 20C5)

= 2 × (20C5 × 20C6)

This is the total number of ways by which a sports team of 11 students can be formed.
In simple words: You need at least 5 from each class. So either pick 5-6 or 6-5. For each case, calculate the ways and add them together.

Exam Tip: With constraint problems involving "at least X," list all valid distributions explicitly. Here: (5,6) and (6,5) are the only two options that total 11 and meet the "at least 5" requirement.

Question 7. From 4 officers and 8 clerks, in how many ways can 6 be chosen (i) to include exactly one officer, (ii) to include at least one officer?
Answer: We need to select 6 people from 4 officers and 8 clerks with certain restrictions.

(i) To include exactly one officer:

One officer is chosen from 4 in 4C1 ways.

The remaining 5 people must be chosen from 8 clerks in 8C5 ways.

Using the multiplication principle:

Total ways = 4C1 × 8C5

(ii) To include at least one officer:

This condition has multiple subcases:

(a) One officer and 5 clerks

(b) Two officers and 4 clerks

(c) Three officers and 3 clerks

(d) Four officers and 2 clerks

Alternatively, we can use: Total selections with at least one officer = Total possible selections - Selections with only clerks

Total ways to choose 6 from 12 = 12C6

Ways to choose 6 clerks (no officers) = 8C6

Total ways = 12C6 - 8C6
In simple words: For (i), pick 1 officer from 4 and 5 clerks from 8 - multiply the ways. For (ii), it's easier to calculate all ways to pick 6 people, then subtract the ways to pick only clerks (no officers).

Exam Tip: For "at least one" problems, the complement rule is usually faster: calculate total ways minus the unwanted case (zero officers), rather than adding up all the subcases.

Question 8. A cricket team of 11 players is to be selected from 16 players including 5 bowlers and 2 wicketkeepers. In how many ways can a team be selected so as to consist of exactly 3 bowlers and 1 wicketkeeper?
Answer: We are selecting 11 players from 16, with the constraint that exactly 3 must be bowlers and exactly 1 must be a wicketkeeper.

This means we need 11 - 3 - 1 = 7 batsmen to complete the team.

From the 5 available bowlers, we choose 3 in 5C3 ways.

From the 2 available wicketkeepers, we choose 1 in 2C1 ways.

The remaining 9 players (after removing bowlers and wicketkeepers) are batsmen. From these 9, we choose 7 in 9C7 ways.

Using the multiplication principle:

Total ways = 5C3 × 2C1 × 9C7

\( ^5C_3 = \frac{5!}{3! \times 2!} = 10 \)

\( ^2C_1 = 2 \)

\( ^9C_7 = \frac{9!}{7! \times 2!} = 36 \)

Total = 10 × 2 × 36 = 720

The team can be selected in 720 different ways.
In simple words: Select 3 bowlers from 5 (10 ways), 1 wicketkeeper from 2 (2 ways), and 7 batsmen from the remaining 9 players (36 ways). Multiply: 10 × 2 × 36 = 720.

Exam Tip: When the constraint specifies exact numbers of different types, work out the composition first. If 11 total with 3 bowlers and 1 wicketkeeper, then 7 must be batsmen. Use multiplication principle on all three groups.

Question 9. In how many ways can a cricket team be selected from a group of 25 players containing 10 batsmen, 8 bowlers, 5 all-rounders and 2 wicketkeepers, assuming that the team of 11 players requires 5 batsmen, 3 all-rounders, 2 bowlers and 1 wicketkeeper?
Answer: We need to select exactly 11 players with a specific distribution from the 25 players available.

Choose 5 batsmen from 10: 10C5

Choose 3 all-rounders from 5: 5C3

Choose 2 bowlers from 8: 8C2

Choose 1 wicketkeeper from 2: 2C1

Using the multiplication principle:

Total ways = 10C5 × 5C3 × 8C2 × 2C1

\( ^{10}C_5 = 252 \)

\( ^5C_3 = 10 \)

\( ^8C_2 = 28 \)

\( ^2C_1 = 2 \)

Total = 252 × 10 × 28 × 2 = 141,120

A cricket team can be selected in 141,120 different ways.
In simple words: Pick 5 batsmen, 3 all-rounders, 2 bowlers, and 1 wicketkeeper using the combination formula for each group. Multiply all four results together.

Exam Tip: Write out the exact requirements first: 5+3+2+1 = 11 total. Calculate each group's combinations independently, then multiply all of them using the multiplication principle.

Question 10. A question paper has two parts, part A and part B, each containing 10 questions. If the student has to choose 8 from part A and 5 from part B, in how many ways can he choose the questions?
Answer: The question paper is divided into two sections - each section has 10 questions. The student must select 8 from part A and 5 from part B.

Number of ways to choose 8 questions from 10 in part A: 10C8

Number of ways to choose 5 questions from 10 in part B: 10C5

Using the multiplication principle:

Total ways = 10C8 × 10C5

\( ^{10}C_8 = \frac{10!}{8! \times 2!} = 45 \)

\( ^{10}C_5 = \frac{10!}{5! \times 5!} = 252 \)

Total = 45 × 252 = 11,340

The student can select questions in 11,340 different ways.
In simple words: From part A, pick 8 out of 10 (45 ways). From part B, pick 5 out of 10 (252 ways). The total number of ways to attempt the paper is 45 × 252 = 11,340.

Exam Tip: Notice that 10C8 = 10C2. Using this symmetry property of combinations can speed up calculations: choosing 8 to include is the same as choosing 2 to exclude.

Question 11. In an examination, a student has to answer 4 questions out of 5. Questions 1 and 2 are compulsory. Find the number of ways in which the student can make a choice.
Answer: The student must answer 4 questions total and is forced to answer questions 1 and 2. This means the student only needs to pick 2 more questions from the remaining 3 questions.

Number of ways to choose 2 questions from the remaining 3: 3C2

Applying nCr = \( \frac{n!}{r!(n-r)!} \):

\( ^3C_2 = \frac{3!}{2! \times 1!} = 3 \)

The student can make a choice in 3 different ways.
In simple words: 2 questions are mandatory. You pick 2 more from the 3 remaining questions. That can be done in 3 ways.

Exam Tip: When items are compulsory, always reduce the problem. Don't calculate 5C4 (choosing 4 from 5). Instead, recognize that 2 are fixed and calculate 3C2 (choosing 2 from the other 3).

Question 12. In an examination, a student has to answer 10 questions, choosing at least 4 from each of part A and part B. If there are 6 questions in part A and 7 in part B, in how many ways can these questions be chosen?
Answer: A total of 13 questions exist, and the student must answer 10. With the constraint of at least 4 from each part, the possible distributions are:

**Case 1:** 6 questions from part A and 4 from part B

**Case 2:** 5 questions from part A and 5 from part B

**Case 3:** 4 questions from part A and 6 from part B

Total ways = (6C6 × 7C4) + (6C5 × 7C5) + (6C4 × 7C6)

\( ^6C_6 = 1 \)

\( ^7C_4 = 35 \)

\( ^6C_5 = 6 \)

\( ^7C_5 = 21 \)

\( ^6C_4 = 15 \)

\( ^7C_6 = 7 \)

Total = (1 × 35) + (6 × 21) + (15 × 7) = 35 + 126 + 105 = 266

The questions can be chosen in 266 different ways.
In simple words: You need 10 total with at least 4 from each part (A has 6, B has 7). The valid splits are 6-4, 5-5, and 4-6. Calculate each case separately, then add them up.

Exam Tip: List all valid distributions that satisfy the constraint before calculating. Here, starting from maximum in A and moving down (6-4, then 5-5, then 4-6) ensures you don't miss any case.

Question 13. In an examination, a candidate is required to answer 7 questions out of 12, which are divided into two groups, each containing 6 questions. One cannot attempt more than 5 questions from either group. In how many ways can he choose these questions?
Answer: There are 12 total questions split into two groups of 6 each. The candidate must choose 7 questions but cannot pick more than 5 from any single group.

The valid distributions are:

**Case 1:** 3 questions from group A and 4 from group B

**Case 2:** 4 questions from group A and 3 from group B

**Case 3:** 5 questions from group A and 2 from group B

**Case 4:** 2 questions from group A and 5 from group B

Total ways = (6C3 × 6C4) + (6C4 × 6C3) + (6C5 × 6C2) + (6C2 × 6C5)

\( ^6C_3 = 20 \)

\( ^6C_4 = 15 \)

\( ^6C_5 = 6 \)

\( ^6C_2 = 15 \)

Total = (20 × 15) + (15 × 20) + (6 × 15) + (15 × 6) = 300 + 300 + 90 + 90 = 780

The candidate can choose the questions in 780 different ways.
In simple words: Pick 7 total, but no more than 5 from either group. Valid splits are 3-4, 4-3, 5-2, and 2-5. Calculate each and add them.

Exam Tip: With upper limit constraints ("cannot attempt more than 5"), explicitly list which distributions are NOT allowed. Here, 6-1, 1-6, 7-0, and 0-7 all violate the rule, so exclude them.

Question 14. Out of 6 teachers and 8 students, a committee of 11 is being formed. In how many ways can this be done, if the committee contains (i) exactly 4 teachers? (ii) at least 4 teachers?
Answer: A committee of 11 is being formed from 6 teachers and 8 students (total 14 people).

(i) Forming a committee with exactly 4 teachers:

Choose 4 teachers from 6 in 6C4 ways.

The remaining 7 members must come from 8 students, so choose 7 from 8 in 8C7 ways.

Total ways = 6C4 × 8C7

\( ^6C_4 = 15 \)

\( ^8C_7 = 8 \)

Total = 15 × 8 = 120

(ii) The number of ways with at least 4 teachers:

This includes three cases:

**Case 1:** 4 teachers and 7 students = 6C4 × 8C7

**Case 2:** 5 teachers and 6 students = 6C5 × 8C6

**Case 3:** 6 teachers and 5 students = 6C6 × 8C5

Total ways = (6C4 × 8C7) + (6C5 × 8C6) + (6C6 × 8C5)

\( ^6C_4 = 15 \)

\( ^8C_7 = 8 \)

\( ^6C_5 = 6 \)

\( ^8C_6 = 28 \)

\( ^6C_6 = 1 \)

\( ^8C_5 = 56 \)

Total = (15 × 8) + (6 × 28) + (1 × 56) = 120 + 168 + 56 = 344
In simple words: For (i), pick exactly 4 teachers and fill the rest with 7 students. For (ii), you can have 4, 5, or 6 teachers (at least 4). Calculate each case and add them.

Exam Tip: In part (ii), "at least 4 teachers" with 11 committee members means you can have 4, 5, or 6 teachers maximum. You cannot have 7 or more because there are only 6 teachers total.

Question 15. A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of (i) exactly 3 girls? (ii) at least 3 girls? (iii) at most 3 girls?
Answer: A committee of 7 is to be formed from 9 boys and 4 girls.

I. Exactly 3 girls:

If exactly 3 girls are in the committee, then 4 boys must be included. The number of ways they can be selected is:

= 4C3 × 9C4 = 4 × 126 = 504

II. At least 3 girls:

The possible distributions are:

(i) 3 girls and 4 boys

(ii) 4 girls and 3 boys

The number of ways they can be selected is:

= (4C3 × 9C4) + (4C4 × 9C3) = (4 × 126) + (1 × 84) = 504 + 84 = 588

III. At most 3 girls:

This includes four distributions:

(i) 0 girls and 7 boys = 4C0 × 9C7 = 1 × 36 = 36

(ii) 1 girl and 6 boys = 4C1 × 9C6 = 4 × 84 = 336

(iii) 2 girls and 5 boys = 4C2 × 9C5 = 6 × 126 = 756

(iv) 3 girls and 4 boys = 4C3 × 9C4 = 4 × 126 = 504

Total = 36 + 336 + 756 + 504 = 1,632

The committee can be formed in these different ways.
In simple words: For (i), pick exactly 3 girls - the rest are boys. For (ii), you can have 3 or 4 girls. For (iii), you can have 0, 1, 2, or 3 girls (at most 3). Calculate each case and add.

Exam Tip: "At most 3" means you include cases with 0, 1, 2, and 3. Don't forget the edge case of 0 girls. "At least 3" means 3 and above. Be careful not to confuse the two.

Question 16. A committee of three persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women?
Answer: The total number of people is 2 + 3 = 5. We need to form a committee of 3 people.

Total number of ways to form a committee of 3 from 5 people:

\( ^5C_3 = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \)

Now, for committees with 1 man and 2 women:

Ways to select 1 man from 2 = 2C1 = 2

Ways to select 2 women from 3 = 3C2 = 3

Total committees with 1 man and 2 women = 2C1 × 3C2 = 2 × 3 = 6
In simple words: Total committees possible = 10. Out of these, the ones with 1 man and 2 women = 6 (pick 1 man from 2, and 2 women from 3, then multiply).

Exam Tip: When asked "how many of these committees" meet a certain condition, calculate the total first, then calculate only those meeting the condition. Don't assume the second answer divides evenly into the first.

Question 17. A committee of 5 is to be formed out of 6 gents and 4 ladies. In how many ways can this be done, when (i) at least 2 ladies are included? (ii) at most 2 ladies are included?
Answer: A committee of 5 is being formed from 6 gents and 4 ladies (10 people total).

(i) Forming a committee with at least 2 ladies:

The possible distributions are:

(i) 2 ladies and 3 gents

(ii) 3 ladies and 2 gents

(iii) 4 ladies and 1 gent

The number of ways they can be selected:

= (4C2 × 6C3) + (4C3 × 6C2) + (4C4 × 6C1)

\( ^4C_2 = 6 \)

\( ^6C_3 = 20 \)

\( ^4C_3 = 4 \)

\( ^6C_2 = 15 \)

\( ^4C_4 = 1 \)

\( ^6C_1 = 6 \)

Total = (6 × 20) + (4 × 15) + (1 × 6) = 120 + 60 + 6 = 186

(ii) The number of ways with at most 2 ladies:

This includes three distributions:

1. 0 ladies and 5 gents = 4C0 × 6C5 = 1 × 6 = 6

2. 1 lady and 4 gents = 4C1 × 6C4 = 4 × 15 = 60

3. 2 ladies and 3 gents = 4C2 × 6C3 = 6 × 20 = 120

Total ways = 6 + 60 + 120 = 186
In simple words: For (i), you can have 2, 3, or 4 ladies (at least 2). For (ii), you can have 0, 1, or 2 ladies (at most 2). Calculate each case and add the results.

Exam Tip: Notice both answers equal 186 - this is because the complement of "at least 2 ladies" in a 5-person committee from 4 ladies and 6 gents is exactly "at most 2 ladies." This is a useful check on your work.

Question 18. From a class of 14 boys and 10 girls, 10 students are to be chosen for a competition, at least including 4 boys and 4 girls. The 2 girls who won the prizes last year should be included. In how many ways can the selection be made?
Answer: Two girls who won prizes last year must be selected. This leaves us to choose 8 more students from the remaining 14 boys and 8 girls. We must ensure at least 4 boys and at least 2 more girls (since 2 are already chosen, and we need at least 4 girls total).

The possible distributions for the remaining 8 students are:

**(4 boys, 4 girls):** 14C4 × 8C4

\( ^{14}C_4 = 1,001 \)

\( ^8C_4 = 70 \)

Ways = 1,001 × 70 = 70,070

\( ^{14}C_5 = 2,002 \)

\( ^8C_3 = 56 \)

**(5 boys, 3 girls):** 14C5 × 8C3 = 2,002 × 56 = 112,112

\( ^{14}C_6 = 3,003 \)

\( ^8C_2 = 28 \)

**(6 boys, 2 girls):** 14C6 × 8C2 = 3,003 × 28 = 84,084

Total ways = 70,070 + 112,112 + 84,084 = 266,266

The selection can be made in 266,266 different ways.
In simple words: Lock in the 2 prize-winning girls. From the remaining 14 boys and 8 girls, pick enough to make 10 total while respecting the "at least 4 boys and 4 girls" rule. Since 2 girls are fixed, pick from (4,4), (5,3), or (6,2) distributions.

Exam Tip: When certain people are mandatory, fix them first and adjust the remaining requirements. Here, with 2 girls fixed, you only need 2 more girls minimum (for a total of 4), not 4 more.

Question 19. Find the number of 5-card combinations out of a deck of 52 cards if at least one of the five cards has to be king.
Answer: In a deck of 52 cards, there are 4 kings and 48 non-kings. We need to find 5-card combinations where at least one card is a king.

The possible distributions are:

1. 1 king and 4 non-king: 4C1 × 48C4

\( ^4C_1 = 4 \)

\( ^{48}C_4 = 194,580 \)

Ways = 4 × 194,580 = 778,320

2. 2 kings and 3 non-kings: 4C2 × 48C3

\( ^4C_2 = 6 \)

\( ^{48}C_3 = 17,296 \)

Ways = 6 × 17,296 = 103,776

3. 3 kings and 2 non-kings: 4C3 × 48C2

\( ^4C_3 = 4 \)

\( ^{48}C_2 = 1,128 \)

Ways = 4 × 1,128 = 4,512

4. 4 kings and 1 non-king: 4C4 × 48C1

\( ^4C_4 = 1 \)

\( ^{48}C_1 = 48 \)

Ways = 1 × 48 = 48

Total ways = 778,320 + 103,776 + 4,512 + 48 = 886,656
In simple words: You can have 1, 2, 3, or 4 kings in your 5-card hand (at least 1 king). For each case, pick that many kings from the 4 available and fill the rest with non-kings. Add all cases together.

Exam Tip: For "at least one king" problems, it's often faster to use the complement rule: total 5-card hands minus hands with zero kings = 52C5 - 48C5. This single subtraction is quicker than adding four cases.

Question 20. Find the number of diagonals of (i) a hexagon, (ii) a decagon, (iii) a polygon of 18 sides
Answer:
(i) Hexagon (6 sides):

A hexagon has 6 vertices. A diagonal connects two non-adjacent vertices. The total number of line segments connecting any two vertices is 6C2.

\( ^6C_2 = \frac{6 \times 5}{2} = 15 \)

However, 6 of these line segments are the sides of the hexagon (not diagonals).

Number of diagonals = 15 - 6 = 9

(ii) Decagon (10 sides):

A decagon has 10 vertices. Total line segments = 10C2

\( ^{10}C_2 = \frac{10 \times 9}{2} = 45 \)

Number of sides = 10

Number of diagonals = 45 - 10 = 35

(iii) Polygon of 18 sides:

An 18-sided polygon has 18 vertices. Total line segments = 18C2

\( ^{18}C_2 = \frac{18 \times 17}{2} = 153 \)

Number of sides = 18

Number of diagonals = 153 - 18 = 135

General formula: For an n-sided polygon, the number of diagonals = \( ^nC_2 - n = \frac{n(n-1)}{2} - n = \frac{n(n-3)}{2} \)
In simple words: Pick any 2 vertices from n vertices - that's nC2 total segments. Subtract the n sides that form the boundary of the polygon. The rest are diagonals.

Exam Tip: Memorize the formula for diagonals: Number of diagonals = \( \frac{n(n-3)}{2} \) where n is the number of sides. This comes from nC2 (all segments) minus n (the sides). This formula is faster than calculating both and subtracting.

Question 20. Find the number of diagonals in a hexagon, decagon, and a polygon with 18 sides.
Answer: To form a diagonal, we need 2 vertices. When we select any 2 vertices from n vertices, we can form lines using the combination formula \( nC_2 \). However, this includes the sides of the polygon. Since a polygon with n sides has n sides, we must exclude them from our count.
Thus, the number of diagonals in a polygon = \( nC_2 - n \)

(i) Hexagon
N = 6
Number of diagonals = \( 6C_2 - 6 = 15 - 6 = 9 \)

(ii) Decagon
N = 10
Number of diagonals = \( 10C_2 - 10 = 45 - 10 = 35 \)

(iii) Polygon with 18 sides
N = 18
Number of diagonals = \( 18C_2 - 18 = 153 - 18 = 135 \)
In simple words: To count diagonals, first find all possible lines between any two vertices. Then remove the sides because they are not diagonals.

Exam Tip: Always remember the formula: diagonals = \( nC_2 - n \). The key is to subtract the number of sides since they are included in \( nC_2 \) but are not diagonals.

Question 21. How many triangles can be obtained by joining 12 points, four of which are collinear?
Answer: We begin by finding the total number of triangles that can be formed from 12 points without any restrictions.
Total points = 12
Triangles from 12 points = \( 12C_3 = 220 \)

However, 4 of these points are collinear. Points that lie on the same line cannot form a triangle. The number of ways to choose 3 points from these 4 collinear points = \( 4C_3 = 4 \)

Since these 4 combinations do not create actual triangles, we subtract them from our total:
Number of triangles = \( 220 - 4 = 216 \)
In simple words: Start by counting all possible triangles. Then subtract the groups of 3 points that all lie on the same line, since they cannot form a real triangle.

Exam Tip: When points are collinear, they cannot form a triangle. Always identify and subtract the invalid combinations from your total count.

Question 22. How many triangles can be formed in a decagon?
Answer: A decagon has 10 vertices. To form a triangle, we need to choose 3 vertices from these 10 available vertices.

The total number of ways to select 3 vertices from 10 = \( 10C_3 \)

\( 10C_3 = \frac{10!}{3!(10-3)!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120 \)

Total number of triangles = 120
In simple words: A triangle needs exactly 3 corners. Since a decagon has 10 corners, we count how many different ways we can pick 3 of them.

Exam Tip: For polygon triangle-counting problems, use the combination formula \( nC_3 \) where n is the number of vertices. No additional restrictions apply unless points are collinear.

Question 23. How many different selections of 4 books can be made from 10 different books, if (i) there is no restriction? (ii) two particular books are always selected? (iii) two particular books are never selected?
Answer:
(i) When there is no restriction
We need to select 4 books from 10 books without any conditions.
Number of ways = \( 10C_4 = \frac{10!}{4!6!} = 210 \) ways

(ii) When two particular books are always selected
Since 2 books are mandatory, we only need to select 2 more books from the remaining 8 books.
Number of ways = \( 8C_2 = \frac{8!}{2!6!} = 28 \) ways

(iii) When two particular books are never selected
We exclude these 2 books and select all 4 books from the remaining 8 books.
Number of ways = \( 8C_4 = \frac{8!}{4!4!} = 70 \) ways
In simple words: (i) Pick any 4 from 10. (ii) If 2 are forced into the selection, pick the remaining 2 from what is left. (iii) If 2 are forbidden, pick all 4 from the other 8.

Exam Tip: Carefully adjust the total pool and the number of selections needed based on whether items are mandatory or forbidden.

Question 24. How many different products can be obtained by multiplying two or more of the numbers 3, 5, 7, 11 without repetition?
Answer: We need to find products by taking two or more of the four numbers {3, 5, 7, 11} and multiplying them together, without using any number more than once.

The total number of such products equals the number of ways to choose 2 numbers plus the number of ways to choose 3 numbers plus the number of ways to choose all 4 numbers:

= \( 4C_2 + 4C_3 + 4C_4 \)
= 6 + 4 + 1
= 11

These 11 distinct products are:
Two-number products: \( 3 \times 5 = 15 \), \( 3 \times 7 = 21 \), \( 3 \times 11 = 33 \), \( 5 \times 7 = 35 \), \( 5 \times 11 = 55 \), \( 7 \times 11 = 77 \)
Three-number products: \( 3 \times 5 \times 7 = 105 \), \( 3 \times 5 \times 11 = 165 \), \( 3 \times 7 \times 11 = 231 \), \( 5 \times 7 \times 11 = 385 \)
Four-number product: \( 3 \times 5 \times 7 \times 11 = 1155 \)
In simple words: We form products by choosing at least 2 numbers from the 4 available. Count each unique selection group separately, then add up all the ways.

Exam Tip: Break the problem into cases by the number of factors. Use combinations to count selections, not permutations, since multiplication is commutative.

Question 25. Find the number of ways in which a committee of 2 teachers and 3 students can be formed out of 10 teachers and 20 students. In how many of these committees (i) a particular teacher is included? (ii) a particular student is included? (iii) a particular student is excluded?
Answer: We need to form a committee with exactly 2 teachers and 3 students from a pool of 10 teachers and 20 students.

(i) When a particular teacher is included
Since one specific teacher is already in the committee, we need to select 1 more teacher from the remaining 9 teachers, and 3 students from 20 students.
Number of ways = \( 9C_1 \times 20C_3 = 9 \times 1140 = 10260 \) ways

(ii) When a particular student is included
Since one specific student is already in the committee, we need to select 2 teachers from 10 teachers and 2 more students from the remaining 19 students.
Number of ways = \( 10C_2 \times 19C_2 = 45 \times 171 = 7695 \) ways

(iii) When a particular student is excluded
We cannot use this particular student, so we select 2 teachers from 10 teachers and 3 students from the remaining 19 students.
Number of ways = \( 10C_2 \times 19C_3 = 45 \times 969 = 43605 \) ways
In simple words: (i) One teacher is fixed; pick 1 more teacher and 3 students. (ii) One student is fixed; pick 2 teachers and 2 more students. (iii) Exclude one student; pick 2 teachers and 3 from the others.

Exam Tip: When a person is mandatory, reduce the available pool and the number needed. When a person is excluded, reduce only the pool.

Question 26. There are 18 points in a plane of which 5 are collinear. How many straight lines can be formed by joining them?
Answer: A line is formed by joining any two distinct points. If all 18 points were in general position (no three collinear), the total number of lines would be \( 18C_2 \).

\( 18C_2 = \frac{18 \times 17}{2} = 153 \)

However, 5 of these points lie on the same line. When we select any 2 points from these 5 collinear points, they do not form a new line - they all determine the same line. The number of such pairs is \( 5C_2 = 10 \).

These 10 pairs would normally be counted as 10 different lines in our calculation, but they actually represent just 1 line. So we subtract 10 and add back 1:

Number of straight lines = \( 18C_2 - 5C_2 + 1 = 153 - 10 + 1 = 144 \) lines
In simple words: Count all possible lines from any 2 points. Then reduce by removing the overcounting caused by collinear points - we subtract the pairs among collinear points and add back the single line they form.

Exam Tip: For collinear points, use the formula: \( nC_2 - mC_2 + 1 \), where n is total points and m is the number of collinear points.

Exercise 9C

Question 1. Out of 12 consonants and 5 vowels, how many words, each containing 3 consonants and 2 vowels, can be formed?
Answer: To form words, we must first select the letters, then arrange them.

Step 1: Select 3 consonants from 12 consonants = \( 12C_3 \) ways
Step 2: Select 2 vowels from 5 vowels = \( 5C_2 \) ways
Step 3: Arrange these 5 selected letters (3 consonants + 2 vowels) = \( 5! \) ways

Total number of words = \( 12C_3 \times 5C_2 \times 5! \)
= \( 220 \times 10 \times 120 \)
= 264,000 words
In simple words: Choose 3 consonants, choose 2 vowels, then arrange all 5 letters in all possible orders.

Exam Tip: When forming words or arrangements, first use combinations to select letters, then use permutations (factorials) to arrange them.

Question 2. How many words, each of 3 vowels and 2 consonants, can be formed from the letters of the word 'INVOLUTE'?
Answer: First, identify the vowels and consonants in 'INVOLUTE':
Vowels: I, O, U, E (4 vowels)
Consonants: N, V, L, T (4 consonants)

Step 1: Select 3 vowels from 4 vowels = \( 4C_3 \) ways
Step 2: Select 2 consonants from 4 consonants = \( 4C_2 \) ways
Step 3: Arrange these 5 selected letters = \( 5! \) ways

Total number of words = \( 4C_3 \times 4C_2 \times 5! \)
= \( 4 \times 6 \times 120 \)
= 2,880 words
In simple words: Identify and count the vowels and consonants in the word. Then choose 3 vowels and 2 consonants, and arrange them in all possible ways.

Exam Tip: Always first identify which letters in the given word are vowels and which are consonants before applying the combination and permutation formulas.

Question 3. The English alphabet has 21 consonants and 5 vowels. How many words with two different consonants and three different vowels can be formed from the alphabet?
Answer: We need to select and arrange letters from the full English alphabet.

Step 1: Select 2 different consonants from 21 consonants = \( 21C_2 \) ways
Step 2: Select 3 different vowels from 5 vowels = \( 5C_3 \) ways
Step 3: Arrange these 5 selected letters (2 consonants + 3 vowels) = \( 5! \) ways

Total number of words = \( 21C_2 \times 5C_3 \times 5! \)
= \( 210 \times 10 \times 120 \)
= 252,000 words
In simple words: Pick 2 consonants from 21, pick 3 vowels from 5, then arrange all 5 letters in every possible order.

Exam Tip: This is a straightforward combination and permutation problem. Make sure you multiply the selections by the arrangements at the end.

Question 4. In how many ways can 4 girls and 3 boys be seated in a row so that no two boys are together?
Answer: To ensure no two boys sit adjacent, we first seat the girls, then place boys in the gaps between them.

Step 1: Arrange 4 girls in a row
Number of ways = \( 4! = 24 \)

Step 2: Identify positions for boys
When 4 girls sit in a row, there are 5 possible positions for boys (before the first girl, between each pair of girls, and after the last girl): _ G _ G _ G _ G _

Step 3: Choose 3 of these 5 positions for the 3 boys and arrange the boys
We need to select and arrange 3 boys in 5 available positions = \( 5P_3 = 60 \) ways

Total number of arrangements = \( 24 \times 60 = 1,440 \) ways
In simple words: Seat all the girls first. This creates spaces where boys can sit. Place the boys in these spaces so they are never next to each other.

Exam Tip: For "no two of one type together" problems, always seat the other group first, then arrange the restricted group in the available gaps.

Question 5. How many words, with or without meaning, can be formed from the letters of the word 'MONDAY', assuming that no letter is repeated, if (i) 4 letters are used at a time? (ii) All letters are used at a time? (iii) All letters are used, but the first letter is a vowel?
Answer: The word 'MONDAY' contains 6 distinct letters: M, O, N, D, A, Y (with vowels O and A).

(i) When 4 letters are used at a time
We select and arrange 4 letters from 6 available letters.
Number of words = \( 6P_4 = \frac{6!}{(6-4)!} = \frac{6!}{2!} = 6 \times 5 \times 4 \times 3 = 360 \) words

(ii) When all letters are used at a time
We arrange all 6 distinct letters.
Number of words = \( 6! = 720 \) words

(iii) When all letters are used, but the first letter must be a vowel
There are 2 vowels: O and A. We treat each separately:
- If the first position is O, we arrange the remaining 5 letters = \( 5! = 120 \) ways
- If the first position is A, we arrange the remaining 5 letters = \( 5! = 120 \) ways
Total = 120 + 120 = 240 words
In simple words: (i) Pick 4 letters and arrange them. (ii) Arrange all 6 letters. (iii) Lock a vowel in the first spot, then arrange the other 5 freely.

Exam Tip: Use permutations (nPr) when order matters and no repetition is allowed. For constraints on the first position, fix it and arrange the rest.

Exercise 9D

Question 1. If \( 20C_r = 20C_{r-10} \) then find the value of \( 17C_r \).
Answer: We use the property of combinations: if \( nC_r = nC_t \), then either r = t or r + t = n.

Given: \( 20C_r = 20C_{r-10} \)

Applying the property (since r ≠ r - 10):
\( r + (r - 10) = 20 \)
\( 2r - 10 = 20 \)
\( 2r = 30 \)
\( r = 15 \)

Therefore, \( 17C_r = 17C_{15} = \frac{17!}{15! \times 2!} = \frac{17 \times 16}{2} = 136 \)
In simple words: When two combinations of the same number are equal, their subscripts either match or add up to that number. Use this fact to solve for r, then calculate the requested combination.

Exam Tip: Always recall the key property: \( nC_r = nC_t \) means r + t = n (when r ≠ t). This property is essential for solving combination equations.

Question 2. If \( 20C_{r+1} = 20C_{r-10} \) then find the value of \( 10C_r \).
Answer: We use the property: if \( nC_r = nC_t \), then r = t or r + t = n.

Given: \( 20C_{r+1} = 20C_{r-10} \)

Applying the property (if r + 1 = r - 10, we get 1 = -10, which is false):
\( (r + 1) + (r - 10) = 20 \)
\( 2r - 9 = 20 \)
\( 2r = 29 \)
\( r = 14.5 \)

Since r must be a non-negative integer for the combination \( 10C_r \) to be defined, and we obtained r = 14.5 (not an integer), the value of \( 10C_r \) cannot be determined.
In simple words: When solving the equation, we get a non-integer value for r. Since combinations only work with whole numbers, the answer does not exist.

Exam Tip: Always check that your answer for r is a non-negative integer. If it is not, state that the combination is undefined.

Question 3. If \( nC_{r+1} = nC_8 \) then find the value of \( 22C_n \).
Answer: We use the property: if \( nC_r = nC_t \), then r = t or r + t = n.

Given: \( nC_{r+1} = nC_8 \)

From the property, either r + 1 = 8 or (r + 1) + 8 = n.

From the first condition: r + 1 = 8, so r = 7

From the second condition: (r + 1) + 8 = n
Substituting r = 7: 8 + 8 = n, so n = 16

Therefore, \( 22C_n = 22C_{16} = \frac{22!}{16! \times 6!} = 74,613 \)
In simple words: Use the combination property to find both r and n. Then substitute n into the requested combination and calculate.

Exam Tip: When you have an equation like \( nC_{r+1} = nC_8 \), apply both possible conditions from the property to find all unknowns.

Question 4. If \( 35C_{n+7} = 35C_{4n-2} \) then find the value of n.
Answer: We use the property: if \( nC_r = nC_t \), then r = t or r + t = n.

Given: \( 35C_{n+7} = 35C_{4n-2} \)

Case 1: n + 7 = 4n - 2
\( 7 + 2 = 4n - n \)
\( 9 = 3n \)
\( n = 3 \)

Case 2: (n + 7) + (4n - 2) = 35
\( 5n + 5 = 35 \)
\( 5n = 30 \)
\( n = 6 \)

Both values satisfy the conditions, so the value of n is either 3 or 6.
In simple words: Apply both rules from the combination property: the subscripts could be equal, or they could add up to 35.

Exam Tip: When solving combination equations, always check both cases from the property. Both values may be valid solutions.

Question 5. Find the values of (i) \( 200C_{198} \), (ii) \( 76C_0 \), (iii) \( 15C_{15} \).
Answer:
(i) \( 200C_{198} \)
Using the property \( nC_r = nC_{n-r} \), we have \( 200C_{198} = 200C_2 = \frac{200 \times 199}{2 \times 1} = 19,900 \)

(ii) \( 76C_0 \)
By definition, \( nC_0 = 1 \) for all n (since 0! = 1)
Therefore, \( 76C_0 = 1 \)

(iii) \( 15C_{15} \)
By definition, \( nC_n = 1 \) for all n
Therefore, \( 15C_{15} = 1 \)
In simple words: Use standard combination properties: combining all items gives 1, choosing nothing gives 1, and complementary combinations are equal.

Exam Tip: Remember the fundamental properties: \( nC_0 = 1 \), \( nC_n = 1 \), and \( nC_r = nC_{n-r} \).

Question 6. If \( mC_1 = nC_2 \) prove that m = \frac{n(n-1)}{2} \).
Answer: Given: \( mC_1 = nC_2 \)

We expand both combinations:
\( mC_1 = \frac{m!}{1!(m-1)!} = m \)

\( nC_2 = \frac{n!}{2!(n-2)!} = \frac{n(n-1)}{2 \times 1} = \frac{n(n-1)}{2} \)

Since \( mC_1 = nC_2 \):
\( m = \frac{n(n-1)}{2} \)

Hence proved.
In simple words: Write out the formulas for both combinations, set them equal, and simplify to show the relationship between m and n.

Exam Tip: For proof questions, always expand the combination formulas using the definition and simplify step by step.

Question 7. Write the value of \( (5C_1 + 5C_2 + 5C_3 + 5C_4 + 5C_5) \).
Answer: We know that \( 5C_1 + 5C_2 + 5C_3 + 5C_4 + 5C_5 = 2^5 - 5C_0 = 2^5 - 1 = 32 - 1 = 31 \)

Alternatively, we can calculate directly:
\( 5C_1 = 5 \)
\( 5C_2 = 10 \)
\( 5C_3 = 10 \)
\( 5C_4 = 5 \)
\( 5C_5 = 1 \)
Sum = 5 + 10 + 10 + 5 + 1 = 31
In simple words: The sum of all combinations from 1 to n equals 2^n minus the combination for 0. Or simply add them up directly.

Exam Tip: Remember the identity: \( nC_0 + nC_1 + nC_2 + ... + nC_n = 2^n \). Therefore, the sum without \( nC_0 \) equals \( 2^n - 1 \).

Question 8. If \( n+1C_3 = 2(nC_2) \), find the value of n.
Answer: Given: \( n+1C_3 = 2(nC_2) \)

Expanding both sides using the combination formula:
\( \frac{(n+1)!}{3!(n-2)!} = 2 \times \frac{n!}{2!(n-2)!} \)

\( \frac{(n+1) \times n \times (n-1)}{3 \times 2 \times 1} = 2 \times \frac{n \times (n-1)}{2 \times 1} \)

\( \frac{(n+1) \times n \times (n-1)}{6} = n \times (n-1) \)

Dividing both sides by \( n(n-1) \) (assuming n ≥ 2):
\( \frac{n+1}{6} = 1 \)

\( n + 1 = 6 \)
\( n = 5 \)
In simple words: Expand both combinations using their definitions, simplify, and solve for n algebraically.

Exam Tip: When given an equation with combinations, expand them fully and simplify before solving. Always verify your answer.

Question 9. If \( nP_r = 720 \) and \( nC_r = 120 \) then find the value of r.
Answer: We use the relationship between permutations and combinations: \( nP_r = r! \times nC_r \)

Given: \( nP_r = 720 \) and \( nC_r = 120 \)

Substituting into the relationship:
\( 720 = r! \times 120 \)
\( r! = \frac{720}{120} = 6 \)

Since 6 = 3!, we have r = 3
In simple words: Use the formula linking permutations and combinations to find the factorial of r, then identify r.

Exam Tip: Always remember: \( nP_r = r! \times nC_r \). This relationship is useful for connecting permutation and combination problems.

Question 10. If \( (n^2-n)C_2 = (n^2-n)C_4 = 120 \) then find the value of n.
Answer: Using the property: if \( nC_r = nC_t \), then r = t or r + t = n.

Given: \( (n^2-n)C_2 = (n^2-n)C_4 \)

Applying the property (since 2 ≠ 4):
\( 2 + 4 = n^2 - n \)
\( n^2 - n = 6 \)
\( n^2 - n - 6 = 0 \)
\( (n - 3)(n + 2) = 0 \)
\( n = 3 \) or \( n = -2 \)

Since n must be positive, n = 3

We can verify: \( (9 - 3)C_2 = 6C_2 = 15 \)... (Note: The problem statement shows 120, but the calculation gives 15. This suggests checking the original values.)
In simple words: When two combinations of the same number are equal, use the property that their subscripts add up to that number. Solve the resulting equation for n.

Exam Tip: Apply the combination property methodically. Always verify by checking if both combinations equal the given value.

Question 11. How many words are formed by 2 vowels and 3 consonants, taken from 4 vowels and 5 consonants?
Answer: We select vowels and consonants, then arrange them to form words.

Step 1: Select 2 vowels from 4 vowels = \( 4C_2 \) ways
Step 2: Select 3 consonants from 5 consonants = \( 5C_3 \) ways
Step 3: Arrange these 5 selected letters (2 vowels + 3 consonants) = \( 5! \) ways

Total number of words = \( 4C_2 \times 5C_3 \times 5! \)
= \( 6 \times 10 \times 120 \)
= 7,200 words
In simple words: Choose 2 vowels from the available vowels, choose 3 consonants from the available consonants, then arrange all 5 letters in every possible sequence.

Exam Tip: When forming words with restrictions, separate the selection and arrangement steps. Select first using combinations, then arrange using factorials.

Question 12. Find the number of diagonals in an n-sided polygon.
Answer: An n-sided polygon has n vertices. A diagonal connects two non-adjacent vertices.

From each vertex, we can draw lines to (n - 1) other vertices. Of these (n - 1) lines, 2 are sides of the polygon (connecting to adjacent vertices), and the remaining (n - 3) are diagonals.

Since there are n vertices, and each vertex contributes (n - 3) diagonals, we get n(n - 3) total. However, each diagonal is counted twice (once from each endpoint), so we divide by 2.

Number of diagonals = \( \frac{n(n-3)}{2} \)

Alternatively, using combinations: total line segments from n vertices = \( nC_2 = \frac{n(n-1)}{2} \). Subtracting the n sides:
Number of diagonals = \( nC_2 - n = \frac{n(n-1)}{2} - n = \frac{n(n-1) - 2n}{2} = \frac{n(n-3)}{2} \)
In simple words: From each corner, you can draw lines to most other corners. Exclude the two neighboring corners since those form sides, not diagonals. Count from all corners and divide by 2 to avoid double-counting.

Exam Tip: The diagonal formula \( \frac{n(n-3)}{2} \) is fundamental. Derive it from first principles to understand why we divide by 2.

Question 13. Three persons enter a railway compartment having 5 vacant seats. In how many ways can they seat themselves?
Answer: Three distinct persons need to occupy 3 of the 5 available seats. The order in which they sit matters (person A in seat 1 is different from person A in seat 2).

This is a permutation problem where we select and arrange 3 persons in 5 seats.

Number of ways = \( 5P_3 = \frac{5!}{(5-3)!} = \frac{5!}{2!} = 5 \times 4 \times 3 = 60 \) ways
In simple words: The first person has 5 choices of seats, the second person has 4 remaining choices, and the third person has 3 remaining choices.

Exam Tip: For seating problems where people are distinguishable and positions matter, use permutations (nPr), not combinations.

Question 14. There are 12 points in a plane, out of which 3 points are collinear. How many straight lines can be drawn by joining any two of them?
Answer: A line is determined by any 2 distinct points. If all 12 points were in general position, the number of lines would be \( 12C_2 = 66 \).

However, 3 of these points are collinear. These 3 points would normally contribute \( 3C_2 = 3 \) different lines in our count, but since they all lie on the same line, they contribute only 1 line instead of 3.

Number of lines = \( 12C_2 - 3C_2 + 1 = 66 - 3 + 1 = 64 \) lines
In simple words: Start with all possible lines from any 2 points. Subtract the overcounting from collinear points. Add back 1 for the single line they all share.

Exam Tip: For collinear points, the correction formula is: \( nC_2 - mC_2 + 1 \), where n is total points and m is collinear points.

Question 15. In how many ways can committee of 5 be made out of 6 men and 4 women, containing at least 2 women?
Answer: "At least 2 women" means we can have 2, 3, or 4 women in the committee. We calculate each case separately.

Case 1: Committee with 2 women and 3 men
Number of ways = \( 4C_2 \times 6C_3 = 6 \times 20 = 120 \) ways

Case 2: Committee with 3 women and 2 men
Number of ways = \( 4C_3 \times 6C_2 = 4 \times 15 = 60 \) ways

Case 3: Committee with 4 women and 1 man
Number of ways = \( 4C_4 \times 6C_1 = 1 \times 6 = 6 \) ways

Total number of ways = 120 + 60 + 6 = 186 ways
In simple words: Break "at least 2 women" into separate cases: exactly 2 women, exactly 3 women, and exactly 4 women. Calculate each case and add them together.

Exam Tip: "At least" problems require you to consider all valid cases. List them systematically and sum the results.

Question 16. There are 13 cricket players, out of which 4 are bowlers. In how many ways can team of 11 be selected from them so as to include at least 3 bowlers?
Answer: With 13 total players (4 bowlers, 9 non-bowlers) and needing 11 players, we want at least 3 bowlers.

Case 1: Select 3 bowlers and 8 non-bowlers
Number of ways = \( 4C_3 \times 9C_8 = 4 \times 9 = 36 \) ways

Case 2: Select 4 bowlers and 7 non-bowlers
Number of ways = \( 4C_4 \times 9C_7 = 1 \times 36 = 36 \) ways

Total number of ways = 36 + 36 = 72 ways
In simple words: Since we need 11 of 13 players, we exclude exactly 2. If at least 3 must be bowlers, we can exclude at most 1 bowler. This gives us two scenarios: exclude 1 bowler and 1 non-bowler, or exclude 0 bowlers and 2 non-bowlers.

Exam Tip: Sometimes it is easier to think in terms of exclusion rather than inclusion, especially when selecting a large subset.

Question 17. How many different committees of 5 can be formed from 6 men and 4 women, if each committee consists of 3 men and 2 women?
Answer: Each committee must have exactly 3 men from 6 and exactly 2 women from 4.

Number of ways to select 3 men from 6 = \( 6C_3 = 20 \) ways
Number of ways to select 2 women from 4 = \( 4C_2 = 6 \) ways

Total number of different committees = \( 20 \times 6 = 120 \) ways
In simple words: Choose 3 men from the available men and choose 2 women from the available women. Multiply these two counts together.

Exam Tip: When a committee has a fixed composition (exactly m of type A and exactly n of type B), multiply the selections for each type.

Question 18. How many parallelograms can be formed from a set of 4 parallel lines intersecting another set of 3 parallel lines?
Answer: A parallelogram is formed when we select 2 lines from the first set of parallel lines and 2 lines from the second set of parallel lines. The four intersection points of these four lines form the vertices of the parallelogram.

Number of ways to select 2 lines from 4 parallel lines = \( 4C_2 = 6 \) ways
Number of ways to select 2 lines from 3 parallel lines = \( 3C_2 = 3 \) ways

Total number of parallelograms = \( 6 \times 3 = 18 \) parallelograms
In simple words: Pick 2 lines from one direction and 2 lines from the other direction. Their intersections automatically form a parallelogram.

Exam Tip: Visualize how two lines from one set and two from another set create four intersection points that form a parallelogram. No additional constraints are needed.