Question

The number of bijective functions $f :\{1,3,5$, $7, \ldots \ldots 99\} \rightarrow\{2,4,6,8, \ldots \ldots , 100\}$, such that $f (3) \geq f (9) \geq f (15) \geq f (21) \geq \ldots f (99), $ is _____

• A. ${ }^{50} P _{17}$
• B. ${ }^{50} P _{33}$
• C. $33 ! \times 17 !$
• D. $\frac{50 \text { ! }}{2}$

Answer 1

The Correct Option is B

The given problem asks for the number of bijective functions where specific conditions on the ordering of a subset of function mappings are satisfied. We need to find the number of bijective functions \(f\) from the set \(\{1,3,5,7, \ldots ,99\}\) to \(\{2,4,6, \ldots ,100\}\) such that \(f(3) \geq f(9) \geq f(15) \geq \ldots \geq f(99)\).

Let's break down the steps needed to solve this problem:

Consider the two sets:

• Set A: \(\{1, 3, 5, 7, \ldots, 99\}\). It contains all odd numbers from 1 to 99. The size of this set is 50.
• Set B: \(\{2, 4, 6, \ldots, 100\}\). It contains all even numbers from 2 to 100. The size of this set is also 50.

The question states that the function must be bijective. This means each element from Set A must map to a unique element in Set B, and vice versa.

Additional Constraint: The function is such that \(f(3) \geq f(9) \geq f(15) \geq \ldots \geq f(99)\) for 17 elements (as there are 17 terms: \(f(3), f(9), f(15), \ldots, f(99)\)).

For all the 17 positions mentioned above to be in non-increasing order, they have to be chosen in a way that this order is automatically satisfied. The remaining mapping between the other elements of Set A and Set B can be arbitrary, subject to bijection conditions.

Choose the elements in Set B for these 17 specific positions such that they are already in non-increasing order. This implies that we are effectively choosing any 17 elements (without rearranging them further) and leaving the remaining 33 elements of Set B to map to the rest of Set A.

We need to choose 33 elements from a total of 50 possible elements to map freely to the rest of elements in Set A:

• This scenario is a permutation problem where we find permutations for 33 out of 50 elements.

The number of such permutations is given by the permutation formula \({}^{50}P_{33}\).

Therefore, the correct answer is \({}^{50}P_{33}\).