# Combinatorics in n (n,2)+(n+1,2)=16

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### 1 Answer

You need to use factorial formula of combinations such that:

`C_n^k = (n!)/(k!*(n - k)!)`

`` `C_n^2 = (n!)/(2!*(n - 2)!)`

`C_(n+1)^2 = ((n+1)!)/(2!*(n+ 1- 2)!) =gt C_(n+1)^2 = ((n+1)!)/(2!*(n- 1)!)`

You need to write the equation in factorial form such that:

`(n!)/(2!*(n - 2)!) + ((n+1)!)/(2!*(n- 1)!) = 16`

You need to write `(n+1)! = n!*(n+1)` and `(n - 1)! = (n - 2)!*(n - 1)` such that:

`(n!)/(2!*(n - 2)!) + (n!*(n+1))/(2!*(n - 2)!*(n - 1)) = 16`

You need to factor out `(n!)/(2!*(n-2)!)` such that:

`(n!)/(2!*(n-2)!)*(1 + (n + 1)/(n-1)) = 16`

You may write `n! = (n-2)!*(n-1)*n` .

`((n-2)!*(n-1)*n)/(2!*(n-2)!)*((n - 1+ n + 1)/(n-1)) = 16`

Reducing like terms yields:

`(n/2)*(2n) = 16 =gt n^2 = 16 =gt n = +-4`

You need to consider the positive value only such that n = 4.

**Hence, the solution to the combinatorial equation is n = 4.**