The number of bijective functions $f:\{1,3,5,7, \ldots, 99\} \rightarrow\{2,4,6,8, \ldots .100\}$, such that $f(3) \geq f(9) \geq f(15) \geq f(21) \geq \ldots . . f(99)$, is ____________.
Solution
<p>As function is one-one and onto, out of 50 elements of domain set 17 elements are following restriction $f(3) > f(9) > f(15)\,.......\, > f(99)$</p>
<p>So number of ways $= {}^{50}{C_{17}}\,.\,1\,.\,33!$</p>
<p>$= {}^{50}{P_{33}}$</p>
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
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