What is n >= 2 if C(n,2) = C(n,1) + 2 ?

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.