If $\frac{\mathrm{n} !}{2 !(\mathrm{n}-2) !}$ and $\frac{\mathrm{n} !}{4 !(\mathrm{n}-4) !}$ are in the ratio $2: 1$, then $\mathrm{n}=$

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Aicharya · Accepted Answer

We have $\frac{n !}{2 !(n-2) !} 	imes \frac{4 !(n-4) !}{n !}=\frac{2}{1}$ $	herefore \frac{(4 	imes 3)}{(n-2)(n-3)}=2 \Rightarrow n^2-5 n+6=6$

If $\frac{\mathrm{n} !}{2 !(\mathrm{n}-2) !}$ and $\frac{\mathrm{n} !}{4 !(\mathrm{n}-4) !}$ are in the…