JEE Mathematics Permutation and Combination MCQs Set 14

Answer: (a) 18

Explanation:
1. Fix A and read the seats clockwise; 'on his right' means the next person clockwise.
2. If B is right of A, then B's right is C or D: arrangements A B C _ _ _ and A B D _ _ _, each with 3! orders for the other three: \( 2\times6=12 \).
3. If C is right of A, B's right cannot be C (C is already next to A), so it is D: place the block BD among the remaining four places, giving \( 3!=6 \) arrangements.
4. Total \( 12+6=18 \).

Teacher's Note:
1. Fix one person to break the circular symmetry.
2. Split into cases by who is on A's right.

Answer: (c) 21

Explanation:
1. F1 cannot take seat 1 and F2 cannot take seat 2. The allowed adjacent seat pairs (F1, F2) are (2,1), (2,3), (3,4), (4,3), (4,5), (5,4).
2. The other three friends then fill the three remaining seats with no one in his own seat: \( 2,\ 3,\ 4,\ 4,\ 4,\ 4 \) ways respectively.
3. Total \( 2+3+4+4+4+4=21 \).

Teacher's Note:
1. List the possible seats of F1 and F2 first.
2. Count the placements of the others pair by pair, avoiding their own seats.

Answer: (c) 20

Explanation:
1. Total numbers \( \frac{6!}{2!3!}=60 \).
2. Each digit appears in each place in proportion to its frequency: total digit sum per place \( =\frac{60}{6}(2+3+3+4+4+4)=10\times20=200 \).
3. So \( N=200\times111111=22222200 \) and \( N/1111110=20 \).

Teacher's Note:
1. Each place has the same total digit sum.
2. Place-value sum for 6 digits is 111111.

Answer: (a) 1.12

Explanation:
1. \( P_n=\binom{n-2}{3} \) and \( Q_n=\frac{n}{n-3}\binom{n-3}{3} \).
2. For \( n=10 \): \( P_{10}=56 \), \( Q_{10}=50 \), difference 6.
3. Ratio \( \frac{56}{50}=1.12 \).

Teacher's Note:
1. Try small values of n until the difference is 6.
2. In a circle the first and last persons are also neighbours.

Answer: (d) 16.50

Explanation:
1. We need non-negative solutions of \( x_1+x_2+x_3+x_4=8 \) (the limit 10 is never reached).
2. Number \( N=\binom{11}{3}=165 \), so \( N/10=16.50 \).

Teacher's Note:
1. Stars and bars with 4 subjects.
2. Since 8 is less than 10 there is no upper limit to worry about.

Answer: (c) 15

Explanation:
1. He fails if he fails in at least one subject: any non-empty set of subjects.
2. Number \( 2^4-1=15 \) (equal to \( \binom41+\binom42+\binom43+\binom44 \)).

Teacher's Note:
1. Count non-empty subsets of the four subjects.
2. Subjects are not in order, so use combinations, not permutations.

Answer: (b) \( {}^{24}C_5 \)

Explanation:
1. Number of visits by the teacher: \( \binom{25}{5} \); a particular kid goes whenever he is among the 5: \( \binom{24}{4} \).
2. Difference \( \binom{25}{5}-\binom{24}{4}=\binom{24}{5} \) by Pascal's rule.

Teacher's Note:
1. Pascal's rule: C(n, r) = C(n - 1, r) + C(n - 1, r - 1).
2. The difference counts visits in which the kid is absent.

Answer: (b) 90

Explanation:
1. The common difference \( d \) can be 1 to 9; for each \( d \) the first term can be any of \( 20-2d \) values.
2. Total \( \sum_{d=1}^{9}(20-2d)=180-90=90 \) (equal to \( \frac{n(n-2)}{4} \) for even \( n \)).

Teacher's Note:
1. Fix the common difference and count starting terms.
2. For even n the count is n(n - 2)/4.

Answer: (a) 10

Explanation:
1. A symmetrical arrangement is fixed by the left half, which holds \( m \) white and \( n \) red coins in \( m+n \) places.
2. Number \( \binom{m+n}{m}=\binom52=10 \).

Teacher's Note:
1. The right half is a mirror image of the left.
2. Choose positions of the white coins in one half.

Answer: (d) NAAIG

Explanation:
1. Letters in order: A, A, G, I, N. Words starting with A: \( 4!=24 \) (the remaining letters A, G, I, N are all different); with G: \( \frac{4!}{2!}=12 \); with I: 12. So 48 words come before the first word starting with N.
2. Words starting with N, in order: NAAGI (49th), NAAIG (50th), NAGAI (51st), ...
3. The 50th word is NAAIG.

Teacher's Note:
1. Count the blocks of words starting with each earlier letter.
2. Then list the words beginning with N in dictionary order.

Answer: (b) 30240

Explanation:
1. Total arrangements \( 8!=40320 \).
2. Red and white together: treat as a block, \( 2\cdot7!=10080 \).
3. Required \( 40320-10080=30240 \) (equal to \( 6\cdot7! \)).

Teacher's Note:
1. 'Never together' is the complement of 'together'.
2. The block can be arranged in 2 ways internally.

Answer: (b) 840

Explanation:
1. Choose 4 non-adjacent seats out of 10: \( \binom74=35 \).
2. Seat the 4 persons in them: \( 4!=24 \).
3. Total \( 35\times24=840 \).

Teacher's Note:
1. For 'no two adjacent' choose seats with gaps: C(n - r + 1, r).
2. Then multiply by the arrangements of the persons.

Answer: (a) 456

Explanation:
1. Total committees \( \binom{12}{5}=792 \).
2. Committees with no non-official member: \( \binom85=56 \); with exactly one: \( 4\times\binom84=280 \).
3. Required \( 792-56-280=456 \).

Teacher's Note:
1. Use the complement of 'fewer than two non-official members'.
2. Split the complement into 0 and 1 non-official.

Answer: (d) \( S_n=S_{n-1}+n \)

Explanation:
1. The \( n \)-th cut crosses the previous \( n-1 \) cuts at \( n-1 \) points, so it is divided into \( n \) parts.
2. Each part splits a piece in two, adding \( n \) pieces: \( S_n=S_{n-1}+n \).

Teacher's Note:
1. Count the new pieces created by the last cut.
2. With S_0 = 1 this gives S_n = 1 + n(n + 1)/2.

Answer: (d) 11

Explanation:
1. \( S_n=1+\frac{n(n+1)}{2} \).
2. \( S_{10}=56\lt60 \) and \( S_{11}=67\ge60 \).
3. So the minimum number of cuts is 11.

Teacher's Note:
1. Find the smallest n with S_n at least 60.
2. S_10 = 56 just falls short.

Answer: (b) 150

Explanation:
1. All distributions: \( 3^5=243 \).
2. Subtract those where some person gets none: \( 3\cdot2^5-3\cdot1^5=93 \).
3. Required \( 243-93=150 \).

Teacher's Note:
1. Use inclusion-exclusion for 'each person gets at least one'.
2. Equivalent: S(5,3) x 3! = 25 x 6.

Answer: (a) \( a_{17}=a_{16}+a_{15} \)

Explanation:
1. The first digit is 1. An integer ending in 1 can be preceded by anything: \( b_n=a_{n-1} \). One ending in 0 must have 1 before it: \( c_n=b_{n-1}=a_{n-2} \).
2. So \( a_n=b_n+c_n=a_{n-1}+a_{n-2} \), a Fibonacci-type recurrence, and (a) is true.
3. Then \( c_{17}=c_{16}+c_{15} \) and \( b_{17}=b_{16}+c_{16} \) are true, so (b) and (c) are false; (d) is false because \( c_{17}+b_{16}=2a_{15}\ne a_{17} \).

Teacher's Note:
1. Find how an n-digit number ending in 1 or 0 is built from a shorter one.
2. Recurrences of Fibonacci type hold for all three sequences.

Answer: (b) 8

Explanation:
1. \( a_1=1,\ a_2=2,\ a_3=3,\ a_4=5,\ a_5=8 \) by the recurrence \( a_n=a_{n-1}+a_{n-2} \).
2. Then \( b_6=a_5=8 \).

Teacher's Note:
1. b_n counts numbers whose last digit is 1, so the first n - 1 digits are unrestricted.
2. Build the sequence up to a_5.

Answer: (d) 7

Explanation:
1. Subtract \( 1,2,3,4,5 \) from the numbers: \( m_i=n_i-i \) are non-decreasing non-negative integers with sum \( 20-15=5 \).
2. The partitions of 5 into at most 5 parts number 7.

Teacher's Note:
1. Reduce to partitions of a smaller number.
2. List the partitions of 5: 5, 4+1, 3+2, 3+1+1, 2+2+1, 2+1+1+1, 1+1+1+1+1.

Answer: (a) 5

Explanation:
1. Blue segments: \( n \). Red segments: \( \binom n2-n \).
2. Equal when \( \frac{n(n-1)}{2}=2n \), so \( n-1=4 \) and \( n=5 \).

Teacher's Note:
1. Adjacent pairs on the circle number n.
2. Solve a simple linear equation.

Answer: (c) 53

Explanation:
1. Card 1 is in envelope 2. Consider card 2: if it goes to envelope 1, the remaining 4 cards are deranged: \( D_4=9 \).
2. If card 2 does not go to envelope 1, treat envelope 1 as the 'home' of card 2 and derange 5 objects: \( D_5=44 \).
3. Total \( 9+44=53 \).

Teacher's Note:
1. Same pattern as the seven-card problem with two fixed cards.
2. D_4 = 9 and D_5 = 44.

Answer: (b) 5

Explanation:
1. \( n=6!\times5! \) (the block of girls plus 5 boys, then the girls inside).
2. For \( m \): choose the 4 girls forming the block (\( 5\cdot4! \) ways as an ordered block, i.e. choose the separate girl in 5 ways and arrange the four) and place the separate girl so that she is not adjacent to the block; computing gives \( m=432000 \).
3. So \( \frac mn=\frac{432000}{86400}=5 \).

Teacher's Note:
1. Count the positions of the separate girl relative to the block of four.
2. n = 86400.