If $^{n} C_{8}=\,^{n} C_{2},$ find $^{n} C_{2}.$
If $^{n} C_{8}=\,^{n} C_{2},$ find $^{n} C_{2}.$
It is known that, $^{n} C_{a}=\,^{n} C_{b} \Rightarrow a=b$ or $m=a+b$
Therefore,
$^{n} C_{8}=\,^{n} C_{2} \Rightarrow n=8+2=10$
$\therefore\,^{n} C_{2}=\,^{10} C_{2}=\frac{10 !}{2 !(10-2) !}=\frac{10 !}{2 ! 8 !}=\frac{10 \times 9 \times 8 !}{2 \times 1 \times 8 !}=45$
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- [IIT 1977]