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Question

# If $\frac{\left(2n\right)!}{3!\left(2n-3\right)!}$ and $\frac{n!}{2!\left(n-2\right)!}$ are in the ratio 44 : 3, find n.

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Solution

## $\frac{\left(2n\right)!}{3!\left(2n-3\right)!}:\frac{n!}{2!\left(n-2\right)!}=44:3\phantom{\rule{0ex}{0ex}}⇒\frac{\left(2n\right)!}{3!\left(2n-3\right)!}×\frac{2!\left(n-2\right)!}{n!}=\frac{44}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{\left(2n\right)\left(2n-1\right)\left(2n-2\right)\left[\left(2n-3\right)!\right]}{3\left(2!\right)\left(2n-3\right)!}×\frac{2!\left(n-2\right)!}{n\left(n-1\right)\left[\left(n-2\right)!\right]}=\frac{44}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{\left(2n\right)\left(2n-1\right)\left(2n-2\right)}{3}×\frac{1}{n\left(n-1\right)}=\frac{44}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{\left(2n\right)\left(2n-1\right)\left(2\right)\left(n-1\right)}{3}×\frac{1}{n\left(n-1\right)}=\frac{44}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{4\left(2n-1\right)n\left(n-1\right)}{3}×\frac{1}{n\left(n-1\right)}=\frac{44}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{4\left(2n-1\right)}{3}=\frac{44}{3}\phantom{\rule{0ex}{0ex}}⇒\left(2n-1\right)=11\phantom{\rule{0ex}{0ex}}⇒2n=12\phantom{\rule{0ex}{0ex}}⇒n=6$

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