The number of bijective functions f : {1, 3, | vedclass

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(B) The domain set $A = \{1, 3, 5, \ldots, 99\}$ contains $50$ elements. The codomain set $B = \{2, 4, 6, \ldots, 100\}$ also contains $50$ elements.S

Vedclass · Accepted Answer

$^{50}P_{33}$

Vedclass · Answer

(B) The domain set $A = \{1, 3, 5, \ldots, 99\}$ contains $50$ elements. The codomain set $B = \{2, 4, 6, \ldots, 100\}$ also contains $50$ elements.Since $f$ is a bijective function,it must be a permutation of the $50$ elements.The condition given is $f(3) \geq f(9) \geq f(15) \geq \ldots \geq f(99)$.Since $f$ is bijective,all values must be distinct,so the condition becomes $f(3) > f(9) > f(15) > \ldots > f(99)$.There are $17$ elements in the sequence $3, 9, 15, \ldots, 99$ (since $99 = 3 + (n-1)6 \implies n = 17$).We need to choose $17$ distinct values from the $50$ available values in the codomain for these $17$ inputs,which can be done in $^{50}C_{17}$ ways.Once chosen,there is only $1$ way to arrange these $17$ values in descending order to satisfy the condition.The remaining $50 - 17 = 33$ elements in the domain can be mapped to the remaining $33$ elements in the codomain in $33!$ ways.Thus,the total number of such functions is $^{50}C_{17} 	imes 33! = \frac{50!}{17! 	imes 33!} 	imes 33! = \frac{50!}{17!} = ^{50}P_{33}$.

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

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