Question

plzzz solve it. .... ​

Answer 1

Answer:-.

$\displaystyle\frac{\frac{\left(\left(2n\right)!\right)}{\left(5!\left(2n-3\right)!\right)}}{\frac{\left(n!\right)}{\left(4!\left(n-2\right)!\right)}}=\frac{52}{5}\quad :\quad n=7$

Step-by-step explanation:

$\displaystyle\frac{\frac{\left(\left(2n\right)!\right)}{\left(5!\left(2n-3\right)!\right)}}{\frac{\left(n!\right)}{\left(4!\left(n-2\right)!\right)}}=\frac{52}{5}$

$\displaystyle\black{\mathrm{Simplify\:}\frac{\frac{\left(2n\right)!}{5!\left(2n-3\right)!}}{\frac{n!}{4!\left(n-2\right)!}}:}$

$\displaystyle\frac{\frac{\left(2n\right)!}{5!\left(2n-3\right)!}}{\frac{n!}{4!\left(n-2\right)!}}$

$\displaystyle\gray{\frac{\frac{\left(2n\right)!}{5!\left(2n-3\right)!}}{\frac{n!}{4!\left(n-2\right)!}}}$

$\displaystyle=\frac{\left(2n\right)!\cdot \:4!\left(n-2\right)!}{5!\left(2n-3\right)!n!}$

$\displaystyle\gray{\mathrm{Cancel\:the\:factorials}:\quad \frac{n!}{\left(n+m\right)!}=\frac{1}{\left(n+1\right)\cdot \left(n+2\right)\cdots \left(n+m\right)}}$

$\gray{\displaystyle\frac{4!}{5!}=\frac{1}{5}}$

$\displaystyle=\frac{\left(2n\right)!\left(n-2\right)!}{5n!\left(2n-3\right)!}$

$\displaystyle\gray{\mathrm{Cancel\:the\:factorials}:\quad \frac{n!}{\left(n-m\right)!}=n\cdot \left(n-1\right)\cdots \left(n-m+1\right),\:n>m}$

$\displaystyle\gray{\frac{\left(2n\right)!}{\left(2n-3\right)!}=2n\left(2n-1\right)\left(2n-2\right)}$

$\displaystyle=\frac{2n\left(n-2\right)!\left(2n-1\right)\left(2n-2\right)}{5n!}$

$\displaystyle\gray{\mathrm{Cancel\:the\:factorials}:\quad \frac{n!}{\left(n+m\right)!}=\frac{1}{\left(n+1\right)\cdot \left(n+2\right)\cdots \left(n+m\right)}}$

$\gray{\displaystyle\frac{\left(n-2\right)!}{n!}=\frac{1}{n\left(n-1\right)}}$

$\displaystyle=\frac{2n\left(2n-1\right)\left(2n-2\right)}{5n\left(n-1\right)}$

$\gray{\mathrm{Factor}\:2n\left(2n-1\right)\left(2n-2\right):\quad 4n\left(2n-1\right)\left(n-1\right)}$

$\gray{\displaystyle=\frac{4n\left(2n-1\right)\left(n-1\right)}{5n\left(n-1\right)}}$

$\displaystyle\gray{\mathrm{Cancel\:}\frac{4n\left(2n-1\right)\left(n-1\right)}{5n\left(n-1\right)}:\quad \frac{4\left(2n-1\right)}{5}}$

$\displaystyle=\frac{4\left(2n-1\right)}{5}$

$\displaystyle\frac{4\left(2n-1\right)}{5}=\frac{52}{5}$

$\gray{\mathrm{Multiply\:both\:sides\:by\:}5}$

$\displaystyle\frac{4\left(2n-1\right)}{5}\cdot \:5=\frac{52}{5}\cdot \:5$

$\gray{\mathrm{Simplify}}$

$4\left(2n-1\right)=52$

$\gray{\mathrm{Divide\:both\:sides\:by\:}4}$

$\displaystyle\frac{4\left(2n-1\right)}{4}=\frac{52}{4}$

$\gray{\mathrm{Simplify}}$

$2n-1=13$

$\gray{\mathrm{Add\:}1\mathrm{\:to\:both\:sides}}$

$2n-1+1=13+1$

$\gray{\mathrm{Simplify}}$

$2n=14$

$\gray{\mathrm{Divide\:both\:sides\:by\:}2}$

$\displaystyle\frac{2n}{2}=\frac{14}{2}$

$\gray{\mathrm{Simplifu}}$

$n=7$

Answer 2

Answer:

Answer:-.

$\displaystyle\frac{\frac{\left(\left(2n\right)!\right)}{\left(5!\left(2n-3\right)!\right)}}{\frac{\left(n!\right)}{\left(4!\left(n-2\right)!\right)}}=\frac{52}{5}\quad :\quad n=7$

Step-by-step explanation:

$\displaystyle\frac{\frac{\left(\left(2n\right)!\right)}{\left(5!\left(2n-3\right)!\right)}}{\frac{\left(n!\right)}{\left(4!\left(n-2\right)!\right)}}=\frac{52}{5}$

$\displaystyle\black{\mathrm{Simplify\:}\frac{\frac{\left(2n\right)!}{5!\left(2n-3\right)!}}{\frac{n!}{4!\left(n-2\right)!}}:}$

$\displaystyle\frac{\frac{\left(2n\right)!}{5!\left(2n-3\right)!}}{\frac{n!}{4!\left(n-2\right)!}}$

$\displaystyle\gray{\frac{\frac{\left(2n\right)!}{5!\left(2n-3\right)!}}{\frac{n!}{4!\left(n-2\right)!}}}$

$\displaystyle=\frac{\left(2n\right)!\cdot \:4!\left(n-2\right)!}{5!\left(2n-3\right)!n!}$

$\displaystyle\gray{\mathrm{Cancel\:the\:factorials}:\quad \frac{n!}{\left(n+m\right)!}=\frac{1}{\left(n+1\right)\cdot \left(n+2\right)\cdots \left(n+m\right)}}$

$\gray{\displaystyle\frac{4!}{5!}=\frac{1}{5}}$

$\displaystyle=\frac{\left(2n\right)!\left(n-2\right)!}{5n!\left(2n-3\right)!}$

$\displaystyle\gray{\mathrm{Cancel\:the\:factorials}:\quad \frac{n!}{\left(n-m\right)!}=n\cdot \left(n-1\right)\cdots \left(n-m+1\right),\:n>m}$

$\displaystyle\gray{\frac{\left(2n\right)!}{\left(2n-3\right)!}=2n\left(2n-1\right)\left(2n-2\right)}$

$\displaystyle=\frac{2n\left(n-2\right)!\left(2n-1\right)\left(2n-2\right)}{5n!}$

$\displaystyle\gray{\mathrm{Cancel\:the\:factorials}:\quad \frac{n!}{\left(n+m\right)!}=\frac{1}{\left(n+1\right)\cdot \left(n+2\right)\cdots \left(n+m\right)}}$

$\gray{\displaystyle\frac{\left(n-2\right)!}{n!}=\frac{1}{n\left(n-1\right)}}$

$\displaystyle=\frac{2n\left(2n-1\right)\left(2n-2\right)}{5n\left(n-1\right)}$

$\gray{\mathrm{Factor}\:2n\left(2n-1\right)\left(2n-2\right):\quad 4n\left(2n-1\right)\left(n-1\right)}$

$\gray{\displaystyle=\frac{4n\left(2n-1\right)\left(n-1\right)}{5n\left(n-1\right)}}$

$\displaystyle\gray{\mathrm{Cancel\:}\frac{4n\left(2n-1\right)\left(n-1\right)}{5n\left(n-1\right)}:\quad \frac{4\left(2n-1\right)}{5}}$

$\displaystyle=\frac{4\left(2n-1\right)}{5}$

$\displaystyle\frac{4\left(2n-1\right)}{5}=\frac{52}{5}$

$\gray{\mathrm{Multiply\:both\:sides\:by\:}5}$

$\displaystyle\frac{4\left(2n-1\right)}{5}\cdot \:5=\frac{52}{5}\cdot \:5$

$\gray{\mathrm{Simplify}}$

$4\left(2n-1\right)=52$

$\gray{\mathrm{Divide\:both\:sides\:by\:}4}$

$\displaystyle\frac{4\left(2n-1\right)}{4}=\frac{52}{4}$

$\gray{\mathrm{Simplify}}$

$2n-1=13$

$\gray{\mathrm{Add\:}1\mathrm{\:to\:both\:sides}}$

$2n-1+1=13+1$

$\gray{\mathrm{Simplify}}$

$2n=14$

$\gray{\mathrm{Divide\:both\:sides\:by\:}2}$

$\displaystyle\frac{2n}{2}=\frac{14}{2}$

$\gray{\mathrm{Simplify}}$

$n=7$