Let S = {1, 2, 3, 4, 5, 6}. Then the probability that a randomly chosen onto function g from S to S satisfies g(3) = 2g(1) is :

Solution

Maths sir · 4 steps

Onto from S to itself means one-one too, so only the pair (g(1), g(3)) matters here.

1. Onto means bijection. For a function from a finite set to itself, onto and one-one go together. So g is just a permutation of {1, 2, 3, 4, 5, 6}, and there are 6! = 720 of them.
2. Favourable pairs. g(3) = 2g(1) must stay inside S, so g(1) can only be 1, 2 or 3. The pairs (g(1), g(3)) are (1, 2), (2, 4) and (3, 6), and each has two different values, as a permutation needs.
3. Fill the rest. For each pair, the other 4 inputs take the 4 unused values in any order, in 4! = 24 ways. That gives 3 × 24 = 72 favourable functions.

   P = (3 × 4!)/6! = 72/720 = 1/10

4. So. The probability is 1/10, option (a). As a check, (g(1), g(3)) is equally likely to be any of the 6 × 5 = 30 ordered pairs of different values, and 3 of them work: 3/30 = 1/10.

The idea to keep. For onto maps from a finite set to itself, count permutations. When the condition involves only a few inputs, count the allowed values for those inputs and let the rest permute freely.

Answer from the NTA final answer key. Solution by PrepWiser.