JEE Mathematics Permutations and Combinations MCQs Set 22

Answer: (c) 19

Explanation:
1. Check \(n=19\): \({}^{21}C_8=203490\) and \({}^{17}P_4=57120\).
2. \(203490:57120=57:16\) (both equal to 3.5625), and the equation has the unique positive integer root \(n=19\).

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) 15

Explanation:
1. \({}^nC_a={}^nC_b\) with \(a\ne b\) gives \(a+b=n\).
2. \((r+2)+(r-2)=30\) so \(r=15\).

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) 108

Explanation:
1. Each digit is in the unit place in \( 3! = 6 \) numbers.
2. Sum of unit digits = \( 6 x (3 + 4 + 5 + 6) = 6 x 18 = 108 \).

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) 240

Explanation:
1. Choose the other 3 letters from the remaining 5: \( 5C3 = 10 \) ways.
2. Arrange the 4 chosen letters in \( 4! = 24 \) ways: \( 10 x 24 = 240 \).

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) 15

Explanation:
1. The four wicket-balls are chosen from the six balls of the over: \( 6C4 \).
2. \( 6C4 = 15 \).

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) 110

Explanation:
1. The two posts are different, so order matters.
2. \( 11P2 = 11 x 10 = 110 \).

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) \(5\times(9!)^2\)

Explanation:
1. The 10 pearls of one colour can be placed in alternate positions in 9! ways (circular arrangement).
2. The 10 pearls of the other colour then fill the 10 gaps in 10! ways.
3. A necklace can be turned over, so divide by 2: \( 9! x 10!/2 = 9! x 10 x 9!/2 = 5 x (9!)^2 \).

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) 12

Explanation:
1. \({}^nC_2=66\) gives \(n(n-1)=132\).
2. \(n=12\).

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) \(\dfrac{9\cdot9!}{2}\)

Explanation:
1. The first digit can be any of the 9 non-zero digits.
2. The remaining 7 places are filled from the other 9 digits: \( 9P7 = 9!/2! \).
3. Total = \(9\times\dfrac{9!}{2}\).

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) 240

Explanation:
1. Total arrangements of BHARAT (\( A \) repeated) = \( 6!/2! = 360 \).
2. \( B \) and \( H \) together: treat as a block, \( 5!/2! x 2! = 120 \).
3. Never together = \( 360 - 120 = 240 \).

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) 720

Explanation:
1. Unit digit: 1, 3, 5 or 7, i.e. 4 ways.
2. Thousands digit: non-zero, 5 ways. Hundreds and tens digits: 6 ways each.
3. \( 4 x 5 x 6 x 6 = 720 \).

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) 601

Explanation:
1. Alphabetical order of letters: \( A, C, H \), I, \( N, S \).
2. Words starting with \( A, C, H \), I or \( N \): \( 5 x 5! = 600 \).
3. Then SACHIN is the first word beginning with \( S \), so its rank is 601.

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) \(7\cdot{}^6C_4\cdot{}^8C_4\)

Explanation:
1. Arrange the 7 other letters \( M \), I, I, I, I, \( P, P \): \( 7!/(4! 2!) = 105 \) ways, and \( 105 = 7 x 15 = 7 x 6C4 \).
2. These create 8 gaps, from which 4 are chosen for the \( S \): \( 8C4 \).
3. Total = \( 7 x 6C4 x 8C4 = 7350 \).

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) 60

Explanation:
1. The four odd digits 3, 3, 5, 5 go in the 4 even positions: \( 4!/(2! 2!) = 6 \) ways.
2. The digits 2, 2, 8, 8, 8 go in the 5 odd positions: \( 5!/(2! 3!) = 10 \) ways.
3. \( 6 x 10 = 60 \).

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) 72

Explanation:
1. Going: \( 4 x 3 = 12 \) ways.
2. Returning he must avoid the routes already used: 2 routes from \( C \) to \( B \) and 3 from \( B \) to \( A \), i.e. \( 2 x 3 = 6 \) ways.
3. Total \( 12 x 6 = 72 \).

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) 41472

Explanation:
1. Arrange the 4 subject groups: \( 4! = 24 \) ways.
2. Within groups: \( 3! x 4! x 3! x 2! = 6 x 24 x 6 x 2 = 1728 \).
3. Total = \( 24 x 1728 = 41472 \).

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) \(9!\times2\)

Explanation:
1. Treat the President and Secretary as one block: 10 units in a circle, \( (10 - 1)! = 9! \) ways.
2. The two can swap inside the block: 2 ways. Total \( 9! x 2 \).

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) 7200

Explanation:
1. \( 4C2 x 5C3 = 6 x 10 = 60 \) selections.
2. Each can be arranged in \( 5! = 120 \) ways: \( 60 x 120 = 7200 \).

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: (b) 18

Explanation:
1. Choose 2 of the 4 lines and 2 of the 3 lines.
2. \( 4C2 x 3C2 = 6 x 3 = 18 \).

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) 60

Explanation:
1. Without restriction: \( 4^3 = 64 \).
2. A child getting all three prizes: 4 ways.
3. \( 64 - 4 = 60 \).

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) 481

Explanation:
1. Order of letters: \( A, C, H \), I, \( R, T \).
2. Words starting with \( A, C, H \) or I: \( 4 x 5! = 480 \).
3. RACHIT is the first word starting with \( R \), so its rank is 481.

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) 35

Explanation:
1. The 2 compulsory courses are fixed; choose the remaining 3 from the other 7 courses.
2. \( 7C3 = 35 \).

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) 94

Explanation:
1. 2 men and 4 women: \( 4C2 x 6C4 = 6 x 15 = 90 \).
2. 3 men and 6 women: \( 4C3 x 6C6 = 4 \).
3. 4 men would need 8 women, which is impossible. Total \( 90 + 4 = 94 \).

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) 56

Explanation:
1. Each visit is a different group of 3 children.
2. \( 8C3 = 56 \).

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.