JEE Mathematics Permutation and Combination MCQs Set 18

Answer: (d) 120

Explanation:
1. 24 = 2³ × 3. Positive solutions: distribute the three 2's among x, y, z in \( \binom52=10 \) ways and the 3 in 3 ways: 30.
2. The sign pattern must give a positive product, so the number of negative factors is even: 0 or 2, which is \( 1+\binom32=4 \) ways.
3. Total \( 30\times4=120 \).

Teacher's Note:
1. Count positive solutions with stars and bars for each prime.
2. An even number of the three variables must be negative.

Answer: (b) 21

Explanation:
1. Put a = 2x + 1, b = 2y + 1, c = 2z + 1 with x, y, z ≥ 0.
2. Then \( 2(x+y+z)+3=13\Rightarrow x+y+z=5 \), with \( \binom{5+2}{2}=21 \) solutions.

Teacher's Note:
1. Write each odd number as 2k + 1.
2. Count non-negative solutions by stars and bars.

Answer: (b) 70

Explanation:
1. Modulo 4: \( 2^a\equiv2 \) for a = 1 and 0 for a ≥ 2; \( 3^b\equiv3 \) for odd b and 1 for even b; \( 5^c\equiv1 \) always.
2. We need \( 2^a+3^b+1\equiv0\pmod4 \).
3. If a = 1: need 3^b ≡ 1, so b even (2 choices). If a ≥ 2: need 3^b ≡ 3, so b odd (3 choices). Count: a = 1: 1·2·5 = 10; a ∈ {2,..,5}: 4·3·5 = 60. Total 70.

Teacher's Note:
1. Work with residues modulo 4.
2. Treat a = 1 separately, since 2¹ is not a multiple of 4.

Answer: (d) 56

Explanation:
1. Each pair of circles meets in at most 2 points.
2. Number of pairs \( \binom82=28 \), so at most \( 2\times28=56 \) points.

Teacher's Note:
1. Two circles intersect in at most two points.
2. Count pairs.

Answer: (b) 1023

Explanation:
1. Each lamp is either on or off: \( 2^{10} \) states.
2. The hall is illuminated if at least one lamp is on, so exclude the all-off state: \( 2^{10}-1=1023 \).

Teacher's Note:
1. At least one lamp must be on.
2. Subtract the all-off case.

Answer: (a) 224

Explanation:
1. The units digit is even: 4 choices (2, 4, 6, 8).
2. The hundreds digit can be any of the remaining 8 digits and the tens digit any of the remaining 7: \( 4\times8\times7=224 \).

Teacher's Note:
1. Fill the restricted position (units digit) first.
2. Then fill the other places from what is left.

Answer: (c) 27

Explanation:
1. Count solutions of \( x+y+z=11 \) with 1 ≤ x, y, z ≤ 6.
2. Number of positive solutions \( \binom{10}2=45 \); subtract those with some variable ≥ 7: put x = 6 + u (u ≥ 1): u + y + z = 5 has \( \binom42=6 \) solutions, for each of 3 variables: 18.
3. 45 − 18 = 27.

Teacher's Note:
1. Use inclusion-exclusion for the upper limit 6.
2. At most one variable can exceed 6 here.

Answer: (a) 535

Explanation:
1. 37 lines meet in \( \binom{37}2=666 \) pairs, but the 13 lines through A give one point instead of \( \binom{13}2=78 \), and the 11 lines through B give one point instead of \( \binom{11}2=55 \).
2. Total \( 666-78-55+1+1=535 \).

Teacher's Note:
1. Pairs of lines through the same point do not give separate points.
2. Add back the two points A and B.

Answer: (d) 720

Explanation:
1. The first and last digits are even: choose them in \( 3\times2=6 \) ways (even digits 2, 4, 6).
2. The four middle places are filled from the remaining 5 digits: \( 5\times4\times3\times2=120 \).
3. Total \( 6\times120=720 \).

Teacher's Note:
1. Fill the two end places first.
2. Then arrange from the remaining digits.

Answer: (b) 20

Explanation:
1. Interior intersection points of diagonals \( =\binom n4=70 \), so n = 8.
2. Number of diagonals \( =\frac{n(n-3)}2=\frac{8\cdot5}{2}=20 \).

Teacher's Note:
1. Every four vertices determine exactly one interior intersection.
2. C(8, 4) = 70.

Answer: (a) 336

Explanation:
1. x + y + z = 5 has \( \binom72=21 \) solutions.
2. Then u + t = 15 has 16 solutions.
3. Total \( 21\times16=336 \).

Teacher's Note:
1. The second equation fixes the sum of the first three variables.
2. Then u + t = 20 − 5.

Answer: (c) \( ^{n-2}C_3 \)

Explanation:
1. Place the 3 chosen objects and the n − 3 unchosen objects; the chosen ones must lie in separate gaps among the unchosen: there are n − 2 gaps.
2. Choose 3 gaps: \( \binom{n-2}{3} \).

Teacher's Note:
1. Use the gap method: choose gaps between unselected objects.
2. n − 3 objects create n − 2 gaps.