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The number of bijective function f 1 , 3 , 5 , 7 , ⋯ , 99 → 2 , 4 , 6 , 8 , ⋯ , 100 if f 3 > f 5 > f 7 ⋯ > f 99 is

Options

  1. AC 1 50
  2. BC 2 50
  3. C50 ! 2
  4. DC 3 50 × 3 !

Correct answer

A. C 1 50

Step-by-step solution

Given, f 1 , 3 , 5 , 7 , ⋯ , 99 → 2 , 4 , 6 , 8 , ⋯ , 100 The number of elements in domain and codomain is 99 . Now, let us assume f 1 = 100 Now as per diagram we have only 1 way for arranging f 3 > f 5 > f 7 ⋯ > f 99 , Similarly if we choose f 1 = 98 then again we have only 1 way for arranging f 3 > f 5 > f 7 ⋯ > f 99 , So we can see the arrangement f 3 > f 5 > f 7 ⋯ > f 99 depends upon f 1 , So number of ways for choosing f 1 is C 1 50

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