JEE Mathematics Permutation and Combination MCQs Set 15

Answer: (c) at least \( 1000 \)

Explanation:
1. Choose \( 4 \) novels: \( \binom64=15 \); choose the dictionary: \( 3 \).
2. The dictionary is fixed in the middle; arrange the \( 4 \) novels in the other places: \( 4!=24 \).
3. Total \( =15\times3\times24=1080 \), which is at least \( 1000 \).

Teacher's Note:
1. Selection and arrangement are separate steps: the middle position is fixed for the dictionary.

Answer: (c) \( \frac{1}{n+1}\binom{2n}{n} \)

Explanation:
1. Total arrangements: \( \binom{2n}{n} \). Bad arrangements (a prefix with more twos than ones) correspond, by reflection, to arrangements with \( n+1 \) ones and \( n-1 \) twos: \( \binom{2n}{n+1} \).
2. Good arrangements \( =\binom{2n}{n}-\binom{2n}{n+1}=\frac{1}{n+1}\binom{2n}{n} \), the Catalan number.

Teacher's Note:
1. This is the ballot problem. The reflection principle gives the count of bad sequences. Check \( n=3 \): \( 20-15=5 \).

Answer: (a) \( 540 \)

Explanation:
1. Letters appearing once: H, E, I, C, S (\( 5 \) letters). Letters appearing more than once: M, A, T (\( 3 \) letters).
2. Odd positions \( 1,3,5 \): letters taken without repetition from the \( 5 \) single letters: \( 5\cdot4\cdot3=60 \).
3. Even positions \( 2,4 \): letters of M, A, T with repetition allowed: \( 3^2=9 \). Total \( =60\times9=540 \).

Teacher's Note:
1. Classify the letters of MATHEMATICS by whether they occur once or more than once, then fill the odd and even positions separately.

Answer: (a) \( 5 \)

Explanation:
1. All girls together: treat them as one block among \( 5 \) boys: \( 6!\cdot5!=86400 \) ways, so \( n=86400 \).
2. Exactly four girls together, the fifth girl separate: the queue has \( 5 \) boys; choose the girl who is separate (\( 5 \)), arrange the other four in a block (\( 4! \)), place the block and the lone girl in distinct gaps of the boys (\( 6\times5 \) ways), and arrange the boys (\( 5! \)): \( m=5\cdot24\cdot30\cdot120=432000 \).
3. \( \frac mn=\frac{432000}{86400}=5 \).

Teacher's Note:
1. Use the gaps method: arrange the boys first and choose different gaps for the girls' block and the single girl.

Answer: (d) \( 126 \)

Explanation:
1. Circle-circle: \( 2\binom82=56 \). Line-circle: \( 2\times8\times4=64 \). Line-line: \( \binom42=6 \).
2. Total \( =56+64+6=126 \).

Teacher's Note:
1. Two circles meet in at most 2 points, a line and a circle in at most 2, and two lines in at most 1.

Answer: (b) \( 50000 \)

Explanation:
1. Consider the numbers \( 00000 \) to \( 99999 \) as five-digit strings. Each of the \( 5 \) positions contains the digit \( 5 \) in \( 10^4 \) of the strings.
2. So the digit \( 5 \) appears \( 5\times10^4=50000 \) times; leading zeros do not affect the count, and \( 10^5 \) has no \( 5 \).

Teacher's Note:
1. By symmetry each digit appears equally often in each place when numbers are written with leading zeros.

Answer: (d) \( \binom{t+m-1}{m} \)

Explanation:
1. A non-decreasing function is determined by how many elements of \( A \) (in order) map to each element of \( B \), i.e. by non-negative integers \( k_1,\dots,k_t \) with \( k_1+\dots+k_t=m \).
2. The number of such solutions is \( \binom{m+t-1}{t-1}=\binom{t+m-1}{m} \). For example \( m=3 \), \( t=4 \) gives \( 20 \).

Teacher's Note:
1. Think of arranging \( m \) identical items into \( t \) ordered groups. This is a stars and bars count.

Answer: (d) \( 5 \)

Explanation:
1. \( T_n=\binom n3 \), so \( T_{n+1}-T_n=\binom{n+1}{3}-\binom n3=\binom n2 \) (Pascal's rule).
2. \( \binom n2=10 \) gives \( n=5 \).

Teacher's Note:
1. Pascal's identity \( \binom{n+1}{3}=\binom n3+\binom n2 \) simplifies the difference.

Answer: (c) \( 520 \)

Explanation:
1. Choosing \( k \) sides of an \( n \)-gon (sides arranged in a circle) with at least \( g \) unchosen sides between any two chosen ones can be done in \( \frac{n}{n-gk}\binom{n-gk}{k} \) ways.
2. Here \( n=20 \), \( k=3 \), \( g=2 \): \( \frac{20}{14}\binom{14}{3}=\frac{20}{14}\cdot364=520 \).

Teacher's Note:
1. This is a circular selection with minimum gaps. The formula can be checked by computer enumeration, which also gives \( 520 \).

Answer: (c) \( 9 \)

Explanation:
1. Door \( d \) is toggled once for each divisor of \( d \).
2. It ends open only if it has an odd number of divisors, i.e. \( d \) is a perfect square.
3. Squares up to \( 81 \): \( 1,4,9,\dots,81 \), which are \( 9 \) doors.

Teacher's Note:
1. Divisors come in pairs except for the square root of a perfect square, so only perfect squares have an odd number of divisors.

Answer: (c) \( N\le100 \)

Explanation:
1. \( N=\binom{10}{3}-\binom63=120-20=100 \).
2. So \( N=100 \), which satisfies \( N\le100 \).

Teacher's Note:
1. Subtract the collinear triples from all triples. Here \( N \) equals exactly \( 100 \).

Answer: (b) \( 485 \)

Explanation:
1. Let \( X \) invite \( i \) ladies and \( 3-i \) men; then \( Y \) must invite \( 3-i \) ladies and \( i \) men.
2. Number of ways \( =\sum_{i=0}^{3}\binom4i\binom3{3-i}\cdot\binom3{3-i}\binom4i=1+144+324+16 \).
3. Total \( =485 \).

Teacher's Note:
1. Condition on the number of ladies from \( X \)'s side. Both friend groups then follow automatically.

Answer: (b) \( \frac{(n+1)(n-2)}{2} \)

Explanation:
1. Exactly two inversions arise in two ways: (a) two separate adjacent swaps, using non-overlapping adjacent pairs of positions: \( \binom{n-2}{2} \) ways; (b) a cyclic shift of three consecutive entries, giving patterns like \( 2,3,1 \) or \( 3,1,2 \): \( 2(n-2) \) ways.
2. Total \( =\binom{n-2}{2}+2(n-2)=\frac{(n-2)(n-3)+4(n-2)}{2}=\frac{(n-2)(n+1)}{2} \). Check: \( n=3 \) gives \( 2 \) and \( n=4 \) gives \( 5 \).

Teacher's Note:
1. The only ways to have exactly two inversions are two separate adjacent swaps or a 3-cycle on consecutive positions. Verify with \( n=4 \): \( 5 \) permutations.

Answer: (a) \( 185 \)

Explanation:
1. Total choices of \( 3 \) points: \( \binom{12}{3}=220 \).
2. Choices of three collinear points: \( \binom73=35 \).
3. Triangles: \( 220-35=185 \).

Teacher's Note:
1. Subtract the triples which lie on the common line since they do not form triangles.

Answer: (d) \( 1 \)

Explanation:
1. \( xxyy=1100x+11y=11(100x+y)=11\left[99x+(x+y)\right] \). For this to be a square, \( 11 \) must divide \( x+y \), and with \( 1\le x\le9 \), \( 0\le y\le9 \) this means \( x+y=11 \).
2. Then the number is \( 11(99x+11)=121(9x+1) \), so \( 9x+1 \) must be a perfect square. For \( x=1,\dots,9 \): \( 10,19,28,37,46,55,64,73,82 \); only \( 64 \) works, at \( x=7 \), \( y=4 \).
3. The only number is \( 7744=88^2 \), so the count is \( 1 \).

Teacher's Note:
1. Factor out \( 11 \) and use divisibility by \( 11 \) of the digit pattern. Check \( x=1,\dots,9 \) in \( 9x+1 \): only \( x=7 \) gives a square.

Answer: (b) \( 5 \)

Explanation:
1. \( x=10! \).
2. For \( y \): choose the repeated letter (\( 10 \) ways), choose the \( 8 \) other distinct letters from the remaining \( 9 \) (\( \binom98=9 \) ways), and arrange the \( 10 \) letters where one appears twice: \( \frac{10!}{2!} \). So \( y=10\cdot9\cdot\frac{10!}{2} \).
3. \( \frac{y}{9x}=\frac{10\cdot9\cdot10!/2}{9\cdot10!}=5 \).

Teacher's Note:
1. In \( y \) the word uses only \( 9 \) distinct letters, one of which appears twice.

Answer: (a) \( 59 \)

Explanation:
1. Bananas: \( 0 \) to \( 3 \) gives \( 4 \) choices; apples: \( 5 \) choices; oranges: \( 3 \) choices.
2. Total selections \( =4\times5\times3=60 \), which includes the empty selection. Excluding it: \( 59 \).

Teacher's Note:
1. For identical objects, multiply the number of possible counts of each kind. Always subtract the empty selection when at least one item is required.

Answer: (d) \( 3m^2+3m+1 \)

Explanation:
1. A garland can be turned over. The three white flowers split the red ones into three gaps \( x,y,z\ge0 \) with \( x+y+z=6m \). Rotating and turning over the garland can permute the three gaps in every possible way, so garlands correspond to unordered triples \( \{x,y,z\} \).
2. These are the partitions of \( 6m \) into at most \( 3 \) parts. Their number is the integer nearest to \( \frac{(6m+3)^2}{12}=3m^2+3m+\frac34 \), i.e. \( 3m^2+3m+1 \).
3. Check: \( m=1 \) gives \( 7 \) (partitions of \( 6 \) into at most three parts), \( m=2 \) gives \( 19 \).

Teacher's Note:
1. Counting garlands means counting partitions of the red flowers into three gaps. Verify with \( m=1 \): the partitions of \( 6 \) into at most three parts number \( 7 \).

Answer: (a) \( 58 \)

Explanation:
1. Alphabetical letters: A, L, L, M, S. Words starting with A: \( \frac{4!}{2!}=12 \); with L: \( 4!=24 \); with M: \( 12 \); so \( 48 \) words precede those starting with S.
2. Starting with S: SA---: \( \frac{3!}{2!}=3 \); SL---: \( 3!=6 \); then SM---: first is SMALL itself.
3. Position \( =12+24+12+3+6+1=58 \).

Teacher's Note:
1. Count the words that come before SMALL letter by letter, adjusting for the repeated letter L.

Answer: (b) \( 120 \)

Explanation:
1. Choose which three statements are answered correctly: \( \binom{10}{3} \).
2. The other seven are then answered incorrectly in exactly one way each, so the number is \( \binom{10}{3}=120 \).

Teacher's Note:
1. Each statement has exactly one correct answer, so fixing the set of correct statements fixes the whole answer sequence.