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JEE Advanced Mathematics Permutation Combination 2026 JEE Advanced 2026 (Paper 2)

JEE Advanced Mathematics Question (2026) — Solution

Question

A bookshelf contains 6 distinct books of Mathematics and 5 distinct books of Physics. From these 11 books, 6 books are chosen at random. Let X be the absolute value of the difference between the number of Mathematics books chosen and the number of Physics books chosen. If is the mean of the random variable X, then the value of 77 is ___________.

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

Total number of ways to choose 6 books from 11 books is ^ 11 C_ 6 = 462. Let M and P be the number of Mathematics and Physics books chosen respectively. Since 6 books are chosen, M + P = 6. The random variable X is given by X = |M - P| = |2M - 6|. The possible selections of (M, P), the corresponding values of X, and the number of ways are as follows: For M=1, P=5: X = |1 - 5| = 4, Number of ways = ^ 6 C_ 1 ^ 5 C_ 5 = 6 1 = 6 For M=2, P=4: X = |2 - 4| = 2, Number of ways = ^ 6 C_ 2 ^ 5 C_ 4 = 15 5 = 75 For M=3, P=3: X = |3 - 3| = 0, Number of ways = ^ 6 C_ 3 ^ 5 C_ 3 = 20 10 = 200 For M=4, P=2: X = |4 - 2| = 2, Number of ways = ^ 6 C_ 4 ^ 5 C_ 2 = 15 10 = 150 For M=5, P=1: X = |5 - 1| = 4, Number of ways = ^ 6 C_ 5 ^ 5 C_ 1 = 6 5 = 30 For M=6, P=0: X = |6 - 0| = 6, Number of ways = ^ 6 C_ 6 ^ 5 C_ 0 = 1 1 = 1 The mean of the random variable X is given by: = X_i f_i f_i = 4(6) + 2(75) + 0(200) + 2(150) + 4(30) + 6(1) 462 = 24 + 150 + 0 + 300 + 120 + 6 462 = 600 462 = 100 77 Therefore, the value of 77 is: 77 = 77 100 77 = 100 Answer: 100

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