2021/113) prove the following identity: $\binom{n}{k}=\binom{n-2}{k}+2\binom{n-2}{k-1}+\binom{n-2}{k-2}$

We can solve the same in 2 ways. Combinotorics way or alegraic ways

We present here to solve in combinonorics way

From n objects we can choose k objects in   $\binom{n}{k}$ ways

let us group the n objects into (n-2,1,1) ways

for choosing k objects this can be done in 3 ways

k  objects from n-2 objects that is from 1st set 0 from 2nd set and 0 from3rd set in $\binom{n-2}{k}$ ways

k-1  objects from n-2 objects that is from 1st set 1 from 2nd set or  1 from 3rd set in $2 * \binom{n-2}{k-1}$ ways

k-2  objects from n-2 objects that is from 1st set 1 from 2nd set and  1 from 3rd set in $\binom{n-2}{k-2}$ ways

as all above 3 are mutually exclusive so no of ways  =$\binom{n-2}{k} + 2 * \binom{n-2}{k-1} + \binom{n-2}{k-2 }$

in  2 ways we have computed the number of choosing k objects from n obects so they must be same or

$\binom{n}{k} = \binom{n-2}{k} + 2 * \binom{n-2}{k-1} + \binom{n-2}{k-2 }$