Let A={1,2,3,4} and B={1,4,9,16}. Then the number of many-one functions f: A → B such that 1 ∈ f( A) is equal to :

Solution

Maths sir · 3 steps

Count every function whose range contains 1, then throw away the one-one ones.

1. Functions hitting 1. Each of the 4 elements of A can go to any of the 4 elements of B, so there are 4⁴ = 256 functions in all. The ones that miss 1 send every element into {4, 9, 16}, and there are 3⁴ = 81 of them. That leaves 256 - 81 = 175 functions with 1 ∈ f(A).
2. Remove one-one functions. Many-one means not one-one. A and B both have 4 elements, so a one-one function uses all of B and its range always contains 1. There are 4! = 24 such functions, and all of them are among the 175.

   (4⁴ - 3⁴) - 4! = 175 - 24 = 151

3. So. The number of many-one functions with 1 ∈ f(A) is 151, option (a).

The idea to keep. When the range must contain a given element, count the functions that miss it and subtract. Many-one functions are all functions minus the one-one ones.

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