Question
If $^n{C_8}{ = ^n}{C_2}$ . find $^n{C_2}$.

Answer

Here $^n{C_8}{ = ^n}{C_2{}}{ \Rightarrow ^n}C_8{ = ^n}{C_{n - 2}}$ $[{\because ^n}{C_r}{ = ^n}{C_{n - r}}]$
$ \Rightarrow 8 = n - 2[{\because ^n}{C_x}{ = ^n}{C_y} \Rightarrow x = y]$
$ \Rightarrow n = 10$ ${\therefore ^n}{C_2}{ = ^{10}}{C_2} = \frac{{10!}}{{2!8!}} = 45$

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