We have to solve : C(n+2,4) = n^2-1

We know that C(n,k) = n!/k!(n-k)!

C(n+2,4) = n^2-1

=> (n +2)! / 4! * ( n - 2)! = n^2 - 1

=> n*(n -1)(n + 1)((n +2) / 4! = ( n - 1)( n + 1)

=> n(n + 2) = 4!

=> n^2 + 2n = 24

=> n^2 + 2n - 24 = 0

=> n^2 + 6n - 4n - 24 = 0

=> n( n + 6) - 4( n + 6) = 0

=> (n - 4)(n + 6) = 0

n - 4 = 0, gives n = 4

n + 6 = 0, gives n = -6

As the factorial is defined for only a positive number we take n = 4.

**The required value of n = 4**