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 :
Solutions · 1
Official solution
Given that f(k) is a multiple of 3 whenever k is a multiple of 4, we need to consider how to map elements from the domain {1, 2, 3, ..., 20} to the codomain {1, 2, 3, ..., 20} following this rule.
1. We first consider the subset of the domain that consists of multiples of 4: {4, 8, 12, 16, 20}. There are 5 elements in this subset.
2. We then consider the subset of the codomain that consists of multiples of 3: {3, 6, 9, 12, 15, 18}. There are 6 elements in this subset.
3. According to the given condition, each of the 5 multiples of 4 must be mapped to a multiple of 3. This can be done in 6C5⋅5!=6! ways, considering that there are 6 options for each of the 5 multiples of 4 (each choice constitutes a combination), and we then consider the permutations of these 5 choices.
4. The remaining 15 elements in the domain (20 original elements minus the 5 multiples of 4) can be mapped onto the remaining 15 elements in the codomain (20 original elements minus the 6 multiples of 3, plus one multiple of 3 that has been assigned to a multiple of 4). This can be done in 15! ways.
So, combining these two cases, the total number of onto functions f is 6!×15!, which corresponds to option C.