JEE Mathematics Permutations and Combinations MCQs Set 24

Answer: (b) Assertion is true, Reason is false

Explanation:
1. Alphabetical order \( E, F \), I, \( W \) gives WIFE as the last of the 24 words, so the assertion is true.
2. Arranging 4 distinct objects gives \( 4! = 24 \), not \( C(4,4) = 1 \), so the reason is false.

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Answer: (d) Both are true and Reason explains Assertion

Explanation:
1. Put the 10 balls in a row; the 9 gaps between them hold 3 separators to make 4 non-empty groups.
2. Choosing the 3 gap positions from 9 is \( 9C3 \), which is exactly the reason given.

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Answer: (d) Assertion is false, Reason is true

Explanation:
1. First digit 5 ways (1-5), next three digits 6 ways each, the last digit then has exactly 2 choices: \( 5 x 216 x 2 = 2160 \), not 216.
2. The rule on the sum of digits is true.

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Answer: (a) \({}^{n+2}C_r\)

Explanation:
1. Apply Pascal's rule twice: \(({}^nC_r+{}^nC_{r-1})+({}^nC_{r-1}+{}^nC_{r-2})={}^{n+1}C_r+{}^{n+1}C_{r-1}={}^{n+2}C_r\).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Answer: (c) \({}^{n+2}C_{r+1}\)

Explanation:
1. \(({}^nC_{r+1}+{}^nC_r)+({}^nC_r+{}^nC_{r-1})={}^{n+1}C_{r+1}+{}^{n+1}C_r={}^{n+2}C_{r+1}\).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Answer: (d) 1024 ways

Explanation:
1. Each of the 10 questions has 2 possible answers: \( 2^{10} = 1024 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Answer: (a) 28

Explanation:
1. Let the numbers of the three kinds be a, \( b, c \) with \( a + b + c = 6 \); the upper limits 20, 10, 7 are never reached.
2. The number of non-negative solutions is \( 8C2 = 28 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Answer: (b) 17

Explanation:
1. Games among the other \(n-2\) players: \({}^{n-2}C_2\). Games by the two ill players: \( 6 + 6 = 12 \).
2. \({}^{n-2}C_2=105\) gives \(n-2=15\), so \( n = 17 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Answer: (d) 7200

Explanation:
1. Vowels \( O, O, O, A \), I (5 of them) in the 5 even places: \( 5!/3! = 20 \) ways.
2. Consonants \( C, R, R, P, T, N \) in the 6 odd places: \( 6!/2! = 360 \) ways.
3. \( 20 x 360 = 7200 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Answer: (a) 1080

Explanation:
1. Choose the two digits that form pairs: \( 4C2 = 6 \).
2. Remaining two places hold the other two digits once each: \( 6!/(2! 2!) = 180 \) arrangements.
3. \( 6 x 180 = 1080 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Answer: (d) 30

Explanation:
1. All 5-digit strings of 1 and 2: \( 2^5 = 32 \).
2. Remove 11111 and 22222: \( 32 - 2 = 30 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Answer: (c) 150

Explanation:
1. Onto functions from 5 balls to 3 boxes: \( 3^5 - 3 x 2^5 + 3 = 243 - 96 + 3 = 150 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Answer: (b) 266

Explanation:
1. 4 from \( A, 6 \) from \( B \): \( 15 x 7 = 105 \).
2. 5 from \( A, 5 \) from \( B \): \( 6 x 21 = 126 \).
3. 6 from \( A, 4 \) from \( B \): \( 1 x 35 = 35 \). Total \( 105 + 126 + 35 = 266 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Answer: (b) 41

Explanation:
1. Total selections: \( 8C3 = 56 \).
2. Selections with Part II but without Part I: choose 2 others from the 6 remaining: \( 6C2 = 15 \).
3. \( 56 - 15 = 41 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Answer: (c) 13

Explanation:
1. Men's games among themselves \({}^nC_2\); games between men and women \(2n\).
2. \(\dfrac{n(n-1)}2-2n=52\) gives \(n^2-5n-104=0\), so \(n=13\).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Answer: (a) 85

Explanation:
1. Three non-collinear points determine a circle: \( 10C3 - 7C3 = 120 - 35 = 85 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Answer: (c) 252

Explanation:
1. Total 3-digit numbers: 900.
2. Numbers with no digit 5: \( 8 x 9 x 9 = 648 \).
3. \( 900 - 648 = 252 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Answer: (b) 505

Explanation:
1. Total 4-digit numbers using 5 digits with repetition: \( 5^4 = 625 \).
2. All different: \( 5P4 = 120 \).
3. \( 625 - 120 = 505 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Answer: (c) 324

Explanation:
1. Alphabetical order: \( A \), I, \( K, N, R, S \).
2. Before \( K \): \( 2 x 120 = 240 \); after \( K \), before \( R \): \( 3 x 24 = 72 \); then I: \( A \) gives 6; then \( S \): \( A, N \) give 4; then \( N \): \( A \) gives 1.
3. Rank = \( 240 + 72 + 6 + 4 + 1 + 1 = 324 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Answer: (b) \({}^{11}C_6\,(9!)^2\)

Explanation:
1. The side with 4 fixed guests needs 5 more, the other side with 3 fixed needs 6 more, from the 11 remaining: \( 11C5 = 11C6 \).
2. Each side then seats its 9 guests in 9! ways.

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Answer: (c) 385

Explanation:
1. He votes for 1, 2, 3 or 4 candidates: \( 10C1 + 10C2 + 10C3 + 10C4 = 10 + 45 + 120 + 210 = 385 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Answer: (d) 196

Explanation:
1. 4 from the first five and 6 from the other 8: \( 5 x 28 = 140 \).
2. All 5 from the first five and 5 from the other 8: \( 1 x 56 = 56 \).
3. \( 140 + 56 = 196 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.

Answer: (a) \(\dfrac{10!}6\)

Explanation:
1. Among the 3! orders of \( A, B, C \), only one is allowed, so the answer is \( 10!/3! = 10!/6 \).

Teacher's Note:
1. Decide first whether order matters: arrangements use permutations, selections use combinations.
2. Handle restrictions by counting the total and subtracting the bad cases, or by treating 'together' items as one block.