Question

7D%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20X%3D%3F" id="TexFormula1" title=" \frac{(15-X)!}{(13-X)!.2!}=\frac{X!}{(X-2)!.2!} \\ \\ \\ \\ X=?" alt=" \frac{(15-X)!}{(13-X)!.2!}=\frac{X!}{(X-2)!.2!} \\ \\ \\ \\ X=?" align="absmiddle" class="latex-formula">

Answer 1

$\frac{(15-x)!}{(13-x)!\cdot2!}=\frac{x!}{(x-2)!\cdot2!}\\ \frac{(14-x)(15-x)}{2}=\frac{(x-1)x}{2}\\ (14-x)(15-x)=(x-1)x\\ 210-14x-15x+x^2=x^2-x\\ 28x=210\\ x=\frac{210}{28}=7.5$

The factorial is defined for natural numbers and since 7.5 isn't one, this equation has no solutions.

Answer 2

$(15-x)!=[15-(x+2)]!\times[15-(x+1)]\times(15-x)\\\\=(15-x-2)!\times(15-x-1)\times(15-x)\\\\=(13-x)!\times(14-x)\times(15-x)\\\\=(13-x)!\times(210-14x-15x+x^2)\\\\=(13-x)!\times(x^2-29x+210)\\\\therefore:\frac{(15-x)!}{(13-x)!\cdot2!}=\frac{(13-x)!\cdot(x^2-29x+210)}{(13-x)!\cdot2!}=\frac{x^2-29x+210}{2}$


$x!=(x-2)!\times(x-1)\times x=(x-2)!\times(x^2-x)\\\\therefore:\frac{x!}{(x-2)!\cdot2!}=\frac{(x-2)!\cdot(x^2-x)}{(x-2)!\cdot2}=\frac{x^2-x}{2}\\\\======================================\\\\\frac{x^2-29x+210}{2}=\frac{x^2-x}{2}\ \ \ \ \ |multiply\ both\ sides\ by\ 2\\\\x^2-29x+210=x^2-x\\\\x^2-x^2-29x+x=-210\\\\-28x=-210\ \ \ \ \ |divide\ both\ sides\ by\ (-28)\\\\x=7.5\notin\mathbb{Z}$

$This\ equation\ has\ not\ solution\ because\ x\ must\ be\ an\ integer!\\\\x\in\O$