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JEE Main 201912 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
12 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Permutation and Combination

JEE Main 2019 · 12 January · Morning Shift · Mathematics · Permutation and Combination

Let S = \ 1, 2, 3, . , 100\ , then number of non-empty subsets A of...

Let S={1,2,3,….,100}S = \left\{1, 2, 3, \ldots . , 100\right\}, then number of non-empty subsets AA of SS such that the product of elements in AA is even is :

Options

  1. A2100−1{2}^{100} - 1
  2. B250+1{2}^{50} + 1
  3. C250(250−1){2}^{50}\left({2}^{50} - 1\right)
  4. D250−1{2}^{50} - 1

Correct answer

C. 250(250−1){2}^{50}\left({2}^{50} - 1\right)

Step-by-step solution

As we know that the product of numbers is even when at least one of the number must be even.

Hence total subsets of AA in which product of numbers is even = Total subsets –– total subsets in which all the elements are odd

=2100−250=250(250−1)= {2}^{100} - {2}^{50} = {2}^{50}\left({2}^{50} - 1\right).

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