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JEE Main 201911 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2019 · 11 January · Evening Shift · Mathematics · Functions

The number of functions f from \ 1,2,3, , 20\ onto \ 1,2,3, , 20\...

The number of functions ff from {1,2,3,…,20}\{1,2,3, \ldots, 20\} onto {1,2,3,…,20}\{1,2,3, \ldots, 20\} such that f(k)f(k) is a multiple of 3,3, whenever kk is a multiple of 4 is:

Options

  1. A65×(15)!6^{5} \times(15) !
  2. B5!×6!5 ! \times 6 !
  3. C(15)!×6!(15) ! \times 6 !
  4. D56×155^{6} \times 15

Correct answer

C. (15)!×6!(15) ! \times 6 !

Step-by-step solution

Domain and codomain ={1,2,3,…,20}=\{1,2,3, \ldots, 20\}. There are five multiple of 4 as 4,8,12,16 and 20 . and there are 6 multiple of 3 as 3,6,9,12,15,18 . Since, when ever kk is multiple of 4 then f(k)f(k) is multiple of 3 then total number of arrangement =6c5×5!=6!={ }^{6} c_{5} \times 5 !=6 ! Remaining 15 elements can be arranged in 15!15 ! ways. Since, for every input, there is an output ⇒\Rightarrow function f(k)f(k) in onto ∴\therefore \quad Total number of arrangement =15!.6!=15 ! .6 !
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