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Question:

The number of functions f from 1, 2, 3, ..., 20 onto 1, 2, 3, ..., 20 such that f(k) is a multiple of 3 whenever k is a multiple of 4, is:

(15)! × 6!

65 × (15)!

56 × 15

5! × 6!

Solution:

f(x) = 3m (3, 6, 9, 12, 15, 18) for k = 4, 8, 12, 16, 20
6 × 5 × 4 × 3 × 2 ways
For rest numbers 15! ways
Total ways = 6! (15!)