We'll recall the formula for combinations:

C(n,2) = n!/2!*(n-2)!

But n!=1*2*3*........*(n-3)*(n-2)*(n-1)*n

(n-2)!= 1*2*3*........*(n-3)*(n-2)

We notice that we can write n!=(n-2)!* (n-1)*n

C(n,2) = n!/2!*(n-2)!=(n-2)!*(n-1)*n/2!*(n-2)!

We'll simplify and we'll get:

C(n,2) = (n-1)*n/1*2

C(n,1)= n!/1!*(n-1)!= (n-1)!*n/(n-1)!

We'll simplify and we'll get:

C(n,1) = n

We'll re-write the given equation in the equivalent form:

(n-1)*n/2 = n+2

We'll multiply by 2 both sides:

(n-1)*n=2*(n+2)

We'll remove the brackets:

n^2 - n = 2n + 4

We'll move all terms to the left side:

n^2 - n - 2n - 4 = 0

n^2 - 3n + 4 = 0

We'll apply quadratic formula:

n1=[-(-3)+sqrt(9+16)]/2

n1=(3+5)/2

n1=4

n2=[-(-3)-sqrt(9+16)]/2

n2=(3-5)/2

n2=-1

Since the second value for n does not belong to the interval of admissible values, n>=2, we'll reject it.

**The valid value for n is: n = 4.**