Let A={1,3,7,9,11} and B={2,4,5,7,8,10,12}. Then the total number of one-one maps f: A → B, such that f(1)+f(3)=14, is :

Solution

Maths sir · 4 steps

Fix f(1) and f(3) first, because the sum condition only touches them. Then fill the rest freely.

1. Pairs adding to 14. From B, the pairs with sum 14 are 2 + 12, 4 + 10 and 7 + 7. The map is one-one, so f(1) ≠ f(3) and 7 + 7 is ruled out.
2. Order matters. Each valid pair can be assigned in two ways, for example f(1) = 2, f(3) = 12 or f(1) = 12, f(3) = 2. That gives 2 × 2 = 4 choices for (f(1), f(3)).
3. Remaining elements. Now 7, 9 and 11 must go one-one into the 5 unused elements of B. That can be done in 5 × 4 × 3 = 60 ways.

   4 × 60 = 240

4. So. The number of one-one maps is 240, option (c).

The idea to keep. Handle the restricted elements first and count their ordered choices. Then the free elements are just a permutation ⁿPᵣ of whatever is left in the codomain.

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