16) If `nC12=nC8`, find `nC17`, `22Cn`.

Explanation

Given

`n_(C_(12))​=n_(C_8)`
​
or, `n_(C_(n−12))​=n_(C_8)`
​
`n−12=8`

or, `n=20`

Then `n_(C_(17))`
​
=`20_(C_(17))`
​
=`20_(C_(20−17))`
​
...