JEE Mathematics Permutation and Combination MCQs Set 11

Answer: (b) \( \dfrac k{3!} \)

Explanation:
1. Distributing equally among 3 named persons gives \( k=\dfrac{(3n)!}{(n!)^3} \).
2. Dividing into 3 unlabelled groups of equal size removes the \( 3! \) orderings of the groups.
3. The number of divisions is \( \dfrac{(3n)!}{(n!)^3\,3!}=\dfrac k{3!} \).

Teacher's Note:
1. Groups of equal size are indistinguishable, so divide by the number of orderings of the groups.

Answer: (c) 1080

Explanation:
1. The pattern is two digits used twice each and the other two digits used once each.
2. Choose the two repeated digits in \( \binom42=6 \) ways; the other two digits are then used once.
3. Arrange the multiset: \( \dfrac{6!}{2!\,2!}=180 \). Total \( 6\times180=1080 \).

Teacher's Note:
1. Choose which digits are repeated, then arrange the multiset.

Answer: (d) none of these

Explanation:
1. There are 5 vowels and 21 consonants. Two vowels and two consonants can occupy positions in \( \binom42=6 \) ways.
2. Each vowel position has 5 choices and each consonant position has 21 choices.
3. Total \( 6\times5^2\times21^2=66150 \), which equals none of the three given values.

Teacher's Note:
1. Choose which positions hold vowels, then fill each position.

Answer: (a) 62

Explanation:
1. Each ball can go to either box: \( 2^6=64 \) ways.
2. Remove the two cases in which all balls go to one box.
3. The number of ways is \( 64-2=62 \).

Teacher's Note:
1. Count all placements, then subtract those with an empty box.

Answer: (a) 72

Explanation:
1. The letters are B, H, A, R, A, T: five distinct letters with A repeated twice.
2. All distinct: \( {}^5P_3=60 \).
3. Two A's together with one other letter: \( 4\times\dfrac{3!}{2!}=12 \). Total \( 72 \).

Teacher's Note:
1. Split into the cases of using zero or two A's.

Answer: (d) none of these

Explanation:
1. For no carry, each column's two digits must sum to at most 9. Ordered pairs of digits \( (u,v) \) with \( u+v\le9 \): \( 55 \).
2. For the thousands column both numbers must be genuine 4-digit numbers, so \( u,v\ge1 \): \( 36 \) pairs.
3. Total \( 36\times55^3=5989500 \), not any of the options.

Teacher's Note:
1. Count the digit pairs column by column; the leading column cannot use 0.

Answer: (a) \( {}^{n-1}C_2 \)

Explanation:
1. Positive solutions of a sum with \( r \) variables number \( \binom{n-1}{r-1} \) (stars and bars).
2. Here \( r=3 \).
3. The count is \( \binom{n-1}2={}^{n-1}C_2 \).

Teacher's Note:
1. Place two dividers among the n - 1 gaps between n identical units.

Answer: (c) 13

Explanation:
1. In an AP the middle term is the average, so \( 3b=21 \) and \( b=7 \).
2. Write \( a=7-d,\ c=7+d \); both must be natural numbers, so \( -6\le d\le6 \).
3. That gives 13 values of \( d \), i.e. 13 ordered triplets.

Teacher's Note:
1. The sum of three terms in AP is three times the middle term.

Answer: (a) 165

Explanation:
1. Write each number as \( 2t+1 \) with \( t\ge0 \).
2. Then \( t_1+t_2+t_3+t_4=8 \) with non-negative integers.
3. The count is \( \binom{11}3=165 \).

Teacher's Note:
1. Substitute 2t + 1 for each odd number to get a non-negative equation.

Answer: (a) 91

Explanation:
1. Put \( y=y'+1,\ z=z'+2 \) with \( y',z'\ge0 \).
2. Then \( x+y'+z'=12 \).
3. The count is \( \binom{14}2=91 \).

Teacher's Note:
1. Shift each variable by its minimum to get a non-negative equation.

Answer: (b) 56

Explanation:
1. Introduce a slack variable \( d\ge1 \) with \( a+b+c+d=9 \) for positive integers.
2. The number of positive solutions is \( \binom{8}3 \).
3. That equals \( 56 \).

Teacher's Note:
1. A slack variable converts the inequality into an equation.

Answer: (a) 36

Explanation:
1. \( \vec r\cdot\vec a=x+y+z=10 \) with \( x,y,z \) natural numbers.
2. Positive solutions: \( \binom92 \).
3. \( =36 \) positions.

Teacher's Note:
1. The dot product reduces the condition to a sum of three natural numbers.

Answer: (b) (2) and (3)

Explanation:
1. \( P=(n-r)(n-r+1)\cdots(n+r) \), a product of \( 2r+1 \) consecutive integers.
2. A product of \( k \) consecutive integers is divisible by \( k! \), so \( P \) is divisible by \( (2r+1)! \), and therefore by \( (2r-1)! \) as well.
3. It need not be divisible by \( (2r+2)! \): for \( n=2,\ r=1 \), \( P=6 \) is not a multiple of \( 24 \).

Teacher's Note:
1. Recognise P as a product of 2r + 1 consecutive integers.

Answer: (c) 6 or 7

Explanation:
1. \( {}^{n+5}P_{n+1}=\dfrac{(n+5)!}{4!} \) and \( {}^{n+3}P_n=\dfrac{(n+3)!}{3!} \), so the ratio is \( \dfrac{(n+5)(n+4)}4 \).
2. Setting \( \dfrac{(n+5)(n+4)}4=\dfrac{11(n-1)}2 \) gives \( n^2-13n+42=0 \).
3. So \( n=6 \) or \( n=7 \).

Teacher's Note:
1. Express both permutations through factorials and divide.

Answer: (c) 7 or 14

Explanation:
1. The AP condition is \( 2\,{}^nC_5={}^nC_4+{}^nC_6 \).
2. Dividing by \( {}^nC_5 \): \( 2=\dfrac5{n-4}+\dfrac{n-5}6 \), i.e. \( n^2-21n+98=0 \).
3. So \( n=7 \) or \( n=14 \).

Teacher's Note:
1. Divide the AP condition by the middle term using the ratio of consecutive binomial coefficients.

Answer: (c) (1), (2) and (3)

Explanation:
1. A product of \( r \) consecutive integers is \( r!\binom{m}{r} \) for some \( m \), so it is divisible by \( r! \).
2. \( r \) divides \( r! \), and \( \sum_{k=1}^{r-1}k=\dfrac{r(r-1)}2 \) divides \( r! \) because \( \dfrac{r!}{r(r-1)/2}=2(r-2)! \) is an integer.
3. So all three divide the product.

Teacher's Note:
1. Show that r and r(r - 1)/2 both divide r factorial.

Answer: (a) \( {}^{31}C_{20}-{}^{21}C_{10} \)

Explanation:
1. Choosing 10 articles from one bag: \( \sum_{k=21}^{30}\binom k{10} \).
2. By the hockey-stick identity this is \( \binom{31}{11}-\binom{21}{11} \).
3. \( \binom{31}{11}=\binom{31}{20} \) and \( \binom{21}{11}=\binom{21}{10} \); the value is \( 84319599 \).

Teacher's Note:
1. Use the hockey-stick identity to sum binomial coefficients with the same lower index.

Answer: (b) 2

Explanation:
1. With all different: \( {}^nP_{n-1}=n! \).
2. With two identical (n - 1 distinct kinds): both identical chosen gives \( \dfrac{(n-1)!}{2!}\binom{n-2}{n-3} \), exactly one gives \( (n-1)! \); total \( \dfrac{n\,(n-1)!}2=\dfrac{n!}2 \).
3. So \( k=\dfrac{n!}{n!/2}=2 \).

Teacher's Note:
1. Split the second case by whether both identical things are chosen.

Answer: (c) both \( 2\times{}^8C_3 \) and \( {}^{10}C_5-2\times{}^8C_4 \)

Explanation:
1. Either neither of the couple is invited: \( {}^8C_5=56 \), or both are invited: \( {}^8C_3=56 \).
2. Total \( 112=2\times{}^8C_3 \).
3. Also \( {}^{10}C_5-2\times{}^8C_4=252-140=112 \), so both expressions are correct.

Teacher's Note:
1. Treat the couple as invited together or not at all.

Answer: (a) \( {}^nP_m \)

Explanation:
1. The seats are numbered, so they are distinguishable.
2. We choose \( m \) seats and assign the persons: \( {}^nC_m\times m! \).
3. Hence the count is \( {}^nP_m={}^nC_m\,m! \).

Teacher's Note:
1. Numbered seats make the arrangement a plain permutation.

Answer: (a) 54

Explanation:
1. The number of ways to give some number of each type is \( 5\times4\times3=60 \).
2. Subtract the selections worth less than Rs. 1: giving 0, 25, 50 or 75 paisa from the first two types without any Re. 1 coin: \( 25a+50b\lt100 \) has 6 solutions.
3. \( 60-6=54 \).

Teacher's Note:
1. Count all selections and subtract those worth less than one rupee.