Answered

Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Our platform provides a seamless experience for finding precise answers from a network of experienced professionals. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

[tex] \frac{(15-X)!}{(13-X)!.2!}=\frac{X!}{(X-2)!.2!} \\ \\ \\ \\ X=?[/tex]

Sagot :

konrad509
[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.
[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]
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We appreciate your time. Please come back anytime for the latest information and answers to your questions. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.