JEE Mains · Maths · STD 12 - 1. relation and function
The number of bijective functions \(f :\{1,3,5, 7, \ldots \ldots . .99\} \rightarrow\{2,4,6,8, \ldots \ldots, 100\}\), such that \(f(3) \geq f(9) \geq f(15) \geq f(21) \geq \ldots \ldots f(99), \quad\) is
- A \({ }^{50} P _{17}\)
- B \({ }^{50} P _{33}\)
- C \(33 ! \times 17 !\)
- D \(\frac{50 !}{2}\)
Answer & Solution
Correct Answer
(B) \({ }^{50} P _{33}\)
Step-by-step Solution
Detailed explanation
As function is one-one and onto, out of 50 elements of domain set 17 elements are following restriction \(f(3)>f(9)>f(15) \ldots . .>f(99)\) So number of ways \(={ }^{50} C_{17} \cdot 1 \cdot 33\) ! \(={ }^{50} P_{33}\)
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