Let A = 0, 1, 2, 3, 4, 5, 6, 7. Then the number of bijective functions f : A → A such that f(1) + f(2) = 3 - f(3) is equal to

Solution

Maths sir · 4 steps

Move f(3) across: three different values from A must add to 3, and only 0, 1, 2 can do that.

1. Rewrite the condition. The condition is the same as f(1) + f(2) + f(3) = 3. Since f is a bijection, these three values are distinct elements of A.
2. Find the values. Three different numbers from {0, 1, …, 7} add up to at least 0 + 1 + 2 = 3. So the sum equals 3 only when the values are exactly {0, 1, 2}.
3. Count both parts. The values 0, 1, 2 can go to the inputs 1, 2, 3 in 3! ways. The other 5 inputs {0, 4, 5, 6, 7} must then go to the other 5 values {3, 4, 5, 6, 7} one-to-one, in 5! ways.

   3! × 5! = 6 × 120 = 720

4. So. The number of such bijective functions is 720.

The common slip: 120. Counting only the 5! ways for the remaining elements, and forgetting that 0, 1, 2 can be shuffled among f(1), f(2), f(3) in 3! ways, gives 120.

The idea to keep. When a condition ties a few outputs of a bijection together, first fix which values they can take. Then multiply the ways to arrange those values by the ways to arrange everything left over.

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