Question

`((15-X)!)/((13-X)!.2!)=(X!)/((X-2)!.2!) \\ \\ \\ \\ X=?`

Answer 1

`(15-x)!=[15-(x+2)]!*[15-(x+1)]*(15-x)\\\\=(15-x-2)!*(15-x-1)*(15-x)\\\\=(13-x)!*(14-x)*(15-x)\\\\=(13-x)!*(210-14x-15x+x^2)\\\\=(13-x)!*(x^2-29x+210)\\\\therefore:((15-x)!)/((13-x)!\cdot2!)=((13-x)!\cdot(x^2-29x+210))/((13-x)!\cdot2!)=(x^2-29x+210)/(2)`



`x!=(x-2)!*(x-1)* x=(x-2)!*(x^2-x)\\\\therefore:(x!)/((x-2)!\cdot2!)=((x-2)!\cdot(x^2-x))/((x-2)!\cdot2)=(x^2-x)/(2)\\\\======================================\\\\(x^2-29x+210)/(2)=(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\\otin\mathbb{Z}`


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

Answer 2

`((15-x)!)/((13-x)!\cdot2!)=(x!)/((x-2)!\cdot2!)\\ ((14-x)(15-x))/(2)=((x-1)x)/(2)\\ (14-x)(15-x)=(x-1)x\\ 210-14x-15x+x^2=x^2-x\\ 28x=210\\ x=(210)/(28)=7.5`

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