Answer :
[tex] \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[/tex]
The factorial is defined for natural numbers and since 7.5 isn't one, this equation has no solutions.
The factorial is defined for natural numbers and since 7.5 isn't one, this equation has no solutions.
[tex](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}[/tex]
[tex]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}[/tex]
[tex]This\ equation\ has\ not\ solution\ because\ x\ must\ be\ an\ integer!\\\\x\in\O[/tex]
[tex]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}[/tex]
[tex]This\ equation\ has\ not\ solution\ because\ x\ must\ be\ an\ integer!\\\\x\in\O[/tex]
