Question
If ${^\text{n+2}}\text{C}_{\text{8}}:{^\text{n-2}}\text{C}_{\text{4}},=57:16,$ Find n.
✓
Answer
We have,
$\Rightarrow \frac{\frac{\text{n+2}!}{8!(\text{n-6})!}}{\frac{(\text{n-2})!}{(\text{n-6})!}}=\frac{57}{16}$
$\Rightarrow \frac{(\text{n+2})(\text{n+1})(\text{n})(\text{n-1})(\text{n}-2)}{8!(\text{n}-2)!}=\frac{57}{16}$
Cancelling (n - 2) from number and denominator
$\Rightarrow (\text{n}+2)(\text{n}+1)(\text{n})(\text{n}-1)=\frac{57\times7\times6\times5\times4\times3\times1\times16}{16}$
$\Rightarrow (\text{n+2})(\text{n}+1)(\text{n})(\text{n}-1)=21\times20\times19\times18$
Comparing both sides,
$\text{n}=19$
$\Rightarrow \frac{\frac{\text{n+2}!}{8!(\text{n-6})!}}{\frac{(\text{n-2})!}{(\text{n-6})!}}=\frac{57}{16}$
$\Rightarrow \frac{(\text{n+2})(\text{n+1})(\text{n})(\text{n-1})(\text{n}-2)}{8!(\text{n}-2)!}=\frac{57}{16}$
Cancelling (n - 2) from number and denominator
$\Rightarrow (\text{n}+2)(\text{n}+1)(\text{n})(\text{n}-1)=\frac{57\times7\times6\times5\times4\times3\times1\times16}{16}$
$\Rightarrow (\text{n+2})(\text{n}+1)(\text{n})(\text{n}-1)=21\times20\times19\times18$
Comparing both sides,
$\text{n}=19$
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