The number of bijective functions f : {1, 3, 5, 7,…., 99} {2, 4, 6, 8,…., 100}, such that $f\left(3\right)\ge f\left(9\right)\ge f\left(15\right)\ge f\left(21\right)\ge ....\ge f\left(99\right),$ is

<p>f : {1, 3, 5, 7, ….,99} {2, 4, 6, 8, …….100}</p><p><math><mrow><mi>f</mi><mrow><mo> (</mo><mrow><mn>3</mn></mrow><mo>)</mo></mrow><mo>≥</mo><mi>f</mi><mrow><mo> (</mo><mrow><mn>9</mn></mrow><mo>)</mo></mrow><mo>≥</mo><mi>f</mi><mrow><mo> (</mo><mrow><mn>1</mn><mn>5</mn></mrow><mo>)</mo></mrow><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>≥</mo><mi>f</mi><mrow><mo> (</mo><mrow><mn>9</mn><mn>9</mn></mrow><mo>)</mo></mrow></mrow></math></p><p><math><mrow><mn>3</mn><mo>+</mo><mrow><mo> (</mo><mrow><mi>n</mi><mo>−</mo><mn>1</mn></mrow><mo>)</mo></mrow><mn>6</mn><mo>=</mo><mn>9</mn><mn>9</mn><mo>⇒</mo><mi>n</mi><mo>=</mo><mn>1</mn><mn>7</mn></mrow></math></p><p>cases f (3) &gt; f (9) &gt; f (15) ……. &gt; f (99)</p><p><math><mrow><mo>∴</mo></mrow></math>&nbsp;from the set {2, 4, 6, …., 100}</p><p>17 distinct numbers can be selected in 50C17 ways again remaining {1, 5, 7, 11, ….} can map in 33! ways</p><p><math><mrow><mo>∴</mo></mrow></math>&nbsp;total number of such required functions</p><p><math><mrow><msup><mrow><mo>=</mo></mrow><mrow><mn>5</mn><mn>0</mn></mrow></msup><msub><mrow><mi>C</mi></mrow><mrow><mn>1</mn><mn>7</mn></mrow></msub><mo>×</mo><mn>3</mn><mn>3</mn><mo>!</mo></mrow></math></p><p><math><mrow><mo>=</mo><mfrac><mrow><mn>5</mn><mn>0</mn><mo>!</mo></mrow><mrow><mn>3</mn><mn>3</mn><mo>!</mo><mtext> </mtext><mtext> </mtext><mn>1</mn><mn>7</mn><mo>!</mo></mrow></mfrac><mo>×</mo><mn>3</mn><mn>3</mn><mo>!</mo><msup><mrow><mo>=</mo></mrow><mrow><mn>5</mn><mn>0</mn></mrow></msup><msub><mrow><mi>P</mi></mrow><mrow><mn>3</mn><mn>3</mn></mrow></msub></mrow></math></p>

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