JEE Mathematics Permutations and Combinations MCQs Set 21

Answer: (c) 60

Explanation:
1. There are four even positions (2nd, 4th, 6th, 8th) for the odd digits 3, 3, 5, 5: \( 4!/(2!2!) = 6 \) ways.
2. The remaining five positions take 2, 2, 8, 8, 8: \( 5!/(2!3!) = 10 \) ways.
3. Total \( 6 x 10 = 60 \).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (d) 96

Explanation:
1. Sorted letters: \( C, C, H \), I, \( N, O \). The first letter \( C \) has no smaller letter, so no words come before in that position.
2. Second letter: words with a smaller second letter \( C, H \), I or \( N \) (before \( O \)) each have \( 4! = 24 \) arrangements of the other four letters, giving \( 4 x 24 = 96 \) words.
3. After \( C-O \), the remaining letters \( C, H \), I, \( N \) are already in alphabetical order, so COCHIN is the first word of its block. Hence 96 words precede it.

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (a) 5292

Explanation:
1. Let the middle digit be n (n = 4, ..., 9); the other four digits are chosen from 0, ..., \( n-1 \).
2. Sets containing 0: \( C(n-1,3) \) sets, each arranged in \( 4! - 3! = 18 \) ways (0 not first); sets without 0: \( C(n-1,4) \) sets, \( 4! = 24 \) ways. Count for \( n \): \( 18C(n-1,3) + 24C(n-1,4) \).
3. \( n = 4..9 \) give 18, 96, 300, 720, 1470, 2688, whose sum is 5292.

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (b) 20

Explanation:
1. All arrangements: \( 6!/2! = 360 \). \( B \) and \( H \) together: \( 5!/2! x 2 = 120 \). So \( m = 240 \).
2. Words beginning with \( B \) and ending with \( T \): the middle \( H, A, R, A \) can be arranged in \( 4!/2! = 12 \) ways, so \( n = 12 \).
3. \( m/n = 20 \).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (d) 3600

Explanation:
1. For an \( n x n \) wall with \( n \) odd the two diagonals share the centre tile, so the blue tiles number \( 2n - 1 = 121 \), giving \( n = 61 \).
2. Total tiles \( 61^2 = 3721 \), so red tiles = \( 3721 - 121 = 3600 \).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (a) 1106

Explanation:
1. 1-digit: only 5 (0 is not a natural number): 1.
2. 2-digit: last digit 0 gives 9; last digit 5 gives 8 (first digit not 0 or 5): 17.
3. 3-digit: last 0: \( 9 x 8 = 72 \); last 5: \( 8 x 8 = 64 \); total 136.
4. 4-digit: last 0: \( 9 x 8 x 7 = 504 \); last 5: \( 8 x 8 x 7 = 448 \); total 952. Overall \( 1 + 17 + 136 + 952 = 1106 \).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (c) 336

Explanation:
1. Total circular arrangements of 7 people: \( 6! = 720 \).
2. Two Americans adjacent: treat them as a block, \( 5! x 2 = 240 \); similarly 240 for the British. Both pairs adjacent: \( 4! x 2 x 2 = 96 \).
3. Separated: \( 720 - 240 - 240 + 96 = 336 \).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (d) 256

Explanation:
1. He is unsuccessful if he passes in at most 4 of the 9 papers.
2. \( C(9,0) + C(9,1) + C(9,2) + C(9,3) + C(9,4) = 1 + 9 + 36 + 84 + 126 = 256 \).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (a) \({}^{15}C_3\times{}^{24}C_2\)

Explanation:
1. The 4th largest is 25, so exactly 3 chosen numbers are larger (from 26..40: 15 numbers) and 2 are smaller (from 1..24).
2. Ways: \( C(15,3) x C(24,2) \).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (c) 185

Explanation:
1. Let \( w, b, r \) be the numbers of white, blue, red flags used (\( w \) <= \( 4, b \) <= \( 1, r \) <= \( 3, w + b + r = 6 \)).
2. (4,1,1): \( 6!/(4!) = 30 \); (4,0,2): \( 6!/(4!2!) = 15 \); (3,1,2): \( 6!/(3!2!) = 60 \); (3,0,3): \( 6!/(3!3!) = 20 \); (2,1,3): \( 6!/(2!3!) = 60 \).
3. Total \( 30 + 15 + 60 + 20 + 60 = 185 \).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (a) 13

Explanation:
1. Let there be \( n \) men. Games among men: \(2\binom n2=n(n-1)\). Games between men and women: \(2\cdot n\cdot2=4n\).
2. \(n(n-1)-4n=66\) gives \(n^2-5n-66=0\), so \( n = 11 \) and the participants number 13.

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (b) 15

Explanation:
1. The remaining \(n-2\) players played all games among themselves: \(\binom{n-2}2\), and the two ill players played \( 3 + 3 = 6 \) games in all.
2. \(\binom{n-2}2+6=84\) gives \(\binom{n-2}2=78\), so \( n - 2 = 13 \) and \( n = 15 \).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (c) 50

Explanation:
1. Selecting \( r \) non-adjacent vertices from \( n \) in a circle: \(\dfrac n{n-r}\binom{n-r}r\).
2. \(\dfrac{10}7\binom73=\dfrac{10}7\cdot35=50\).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (d) 90

Explanation:
1. The numbers of objects must be different: (0,1,4) or (0,2,3) in some order.
2. (0,1,4): \( 5!/(1!4!) = 5 \) ways \( x 3! \) orders = 30. (0,2,3): \( 5!/(2!3!) = 10 \) ways \( x 3! = 60 \).
3. Total 90.

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (a) 8

Explanation:
1. The middle term is 7. Take common difference \( d \): \( a = 7 - d, c = 7 + d \), with a >= 1 so \( |d| \) <= 6.
2. \( d = -6 \), ..., 6 gives 13 triplets, so \( \lambda = 13 \) and \( \lambda - 5 = 8 \).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (b) 6

Explanation:
1. The 267th permutation has zero-based index 266. Blocks of 120 by the first digit: \( 266 = 2 x 120 + 26 \), so the first digit is 3 (digits 1, 2, 3 ...).
2. Remaining digits 1, 2, 4, 5, 6 in blocks of 24: \( 26 = 1 x 24 + 2 \), so the second digit is 2.
3. Remaining 1, 4, 5, 6 in blocks of 6: \( 2 = 0 x 6 + 2 \), third digit 1. Remaining 4, 5, 6 in blocks of 2: \( 2 = 1 x 2 + 0 \), fourth digit 5, then 4, 6.
4. The number is 321546, whose unit digit is 6.

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (d) 7

Explanation:
1. Let \( a(n) \) be the ways for \( n \) steps: \( a(n) = a(n-1) + a(n-2) \) with \( a(1) = 1, a(2) = 2 \).
2. \( a(3) = 3, a(4) = 5, a(5) = 8, a(6) = 13, a(7) = 21 \), so \( N = 21 \) and \( N/3 = 7 \).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (a) 31

Explanation:
1. The exponent of 2 in 33! is \(\lfloor33/2\rfloor+\lfloor33/4\rfloor+\lfloor33/8\rfloor+\lfloor33/16\rfloor+\lfloor33/32\rfloor=16+8+4+2+1=31\).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (c) \(36\times55^3\)

Explanation:
1. In each of the last three places the digit pair has sum at most 9: \(1+2+\cdots+10=55\) ordered pairs.
2. In the thousands place both digits are at least 1 with sum at most 9: \(8+7+\cdots+1=36\) pairs.
3. Total \(36\times55^3\).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (d) \(\dfrac{(n-1)^2}4\)

Explanation:
1. With \(n=2k+1\) and common difference \(d\), the first term can be chosen in \(n-2d\) ways, for \(d=1,\ldots,k\).
2. \(\sum_{d=1}^k(2k+1-2d)=k(2k+1)-k(k+1)=k^2=\dfrac{(n-1)^2}4\).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (a) 16

Explanation:
1. All triangles: \( C(8,3) = 56 \).
2. Triangles with two polygon sides: 8. Triangles with exactly one polygon side: \( 8 x (8 - 4) = 32 \).
3. Triangles with no polygon side: \( 56 - 8 - 32 = 16 \).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (c) 1

Explanation:
1. Every friend is present at 7 and absent at 7 dinners, so there were 14 dinners in total.
2. Dinners with at least one friend, by inclusion-exclusion: \( 6(7) - 15(5) + 20(4) - 15(3) + 6(2) - 1 = 42 - 75 + 80 - 45 + 12 - 1 = 13 \).
3. Dinners alone: \( 14 - 13 = 1 \).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (a) 7

Explanation:
1. \( 1000 = 2^3 x 5^3 \) and \( 2000 = 2^4 x 5^3 \). Treat each prime separately and multiply.
2. Powers of 2 (x, y, z for a, \( b, c \)): \( \max(x,y) = 3 \) while \( \max(y,z) = \max(z,x) = 4 \), so \( z = 4 \) and \( x, y \) <= 3 with maximum 3: \( 4^2 - 3^2 = 7 \) ways.
3. Powers of 5: all three pairwise maxima equal 3, so at most one exponent is below 3: all equal to 3 (1 way) or exactly one smaller (3 positions \( x 3 \) values = 9 ways): 10 ways.
4. Total \( 7 x 10 = 70 = 10q \), so \( q = 7 \).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (b) 2

Explanation:
1. Alphabetical order of the letters: \( A, C, E, K, L, T \). Words starting with a smaller letter than \( T \): \( 5 x 5! = 600 \).
2. After \( T, A, C \) the remaining letters are \( E, K, L \); the fourth letter \( K \) has the smaller letter \( E \) before it: \( 1 x 2! = 2 \) words. Then the fifth letter \( L \) has \( E \) before it: 1 word. So 603 words come before TACKLE.
3. Rank = \( 604 = 100 x 6 + 4 \), so \( a = 6, b = 4 \) and \( a - b = 2 \).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (b) 5

Explanation:
1. The number of zeros is \(E_5(n)=\lfloor n/5\rfloor+\lfloor n/25\rfloor+\lfloor n/125\rfloor\).
2. \(E_5(110)=22+4=26\) and the value stays 26 until \(n=114\); \(E_5(115)=23+4=27\). So \( n = 110, 111, 112, 113, 114 \).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (c) 11754

Explanation:
1. Possible multiplicities of the digits used: (6), (4, 2), (3, 3), (2, 2, 2).
2. (6): 9 numbers. (4,2): \( 10 x 9 x 6!/(4!2!) = 1350 \) strings, of which one tenth start with 0: 1215. (3,3): \( 45 x 20 = 900 \) strings, minus 90 starting with 0: 810. (2,2,2): \( 120 x 90 = 10800 \) strings, minus 1080 starting with 0: 9720.
3. Total \( 9 + 1215 + 810 + 9720 = 11754 \).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (d) 25

Explanation:
1. This is the Stirling number of the second kind: \(S(5,3)=25\) (splits 3+1+1: 10 ways, 2+2+1: 15 ways).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (c) 677

Explanation:
1. There are \(26\times26=676\) possible pairs of initials.
2. By the pigeonhole principle \(676+1=677\) students guarantee a repeat.

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (b) 450

Explanation:
1. Choose the first two digits freely (\(9\times10=90\) ways); the last digit then has exactly 5 choices that make the digit sum even.
2. \(90\times5=450\).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (d) 54

Explanation:
1. \(140=2^2\cdot5\cdot7\). The exponent of 2 can be distributed in \(\binom42=6\) ways among \( x, y, z \); the 5 and the 7 each go to one of three variables: 3 ways each.
2. \(6\times3\times3=54\).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.

Answer: (a) 120

Explanation:
1. Introduce a slack variable \(w\ge1\): \(x+y+z+w=11\) in positive integers.
2. \(\binom{10}3=120\).

Teacher's Note:
1. Decide first whether order matters and whether repetition is allowed, then count in stages.
2. Check a counting formula on a small case before trusting it for a larger one.