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 :
(1) 65 x (15)!
(2) 5! x 6!
(3) 56x 15
(4) (15)! x 6!
Correct option (4) (15)! × 6!
Explanation:
f(k) = (3, 6, 9, 12, 15, 18)
For k = 4, 8, 12, 16, 20
For these k we have 6.5.4, 3.2 ways = 6!
For other numbers we have 15! ways.
So total = 15!.6!