The number of bijective functions f:{1,3,5,7, …, 99} →{2,4,6,8, … .100}, such that f(3) ≥ f(9) ≥ f(15) ≥ f(21) ≥ … . . f(99), is ____________.

Solution

Maths sir · 4 steps

A bijection makes the chain strict, so pick 17 values and let the order fix itself. Then fill the rest freely.

1. Count the sets. The domain {1, 3, …, 99} and the codomain {2, 4, …, 100} each have 50 elements. The constrained inputs 3, 9, 15, …, 99 go up in steps of 6, so there are (99-3)/6 + 1 = 17 of them.
2. Chain becomes strict. A bijection never gives two inputs the same value, so every ≥ in the chain is really >. Choose any 17 of the 50 values in C(50, 17) ways. Only one arrangement is decreasing, so the 17 constrained inputs can take them in exactly one way.
3. Fill the rest. The other 33 inputs take the 33 values left over, in any order. That gives 33! ways.

   C(50, 17) × 33! = 50!/(17! 33!) × 33! = 50!/17! = ⁵⁰P₃₃

4. So. The number of such bijections is ⁵⁰P₃₃, which is option (b).

The idea to keep. When some positions must be in a fixed order, choose their values with a combination and do not arrange them. Arrange only the free positions. For a bijection, ≥ is the same as >.

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