What is the natural number n if C(2n-3,2)=3?

By definition, C(2n-3,2) =
(2n-3)!/2!*(2n-3-2)!

C(2n-3,2) =
(2n-3)!/2!*(2n-5)!

But (2n-5)! = (2n - 5)!(2n - 4)(2n -
3)

2! = 1*2 = 2

C(2n-3,2) =
(2n - 5)!(2n - 4)(2n - 3)/2*(2n - 5)!

We'll simplify and
we'll get:

C(2n-3,2) = (2n - 4)(2n -
3)/2

Now, we'll solve the
equation:

(2n - 4)(2n - 3)/2 =
3

(2n - 4)(2n - 3) = 6

We'll
remove the brackets:

4n^2 - 6n - 8n + 12 =
6

4n^2 - 14n + 6 = 0

We'll
divide by 2:

2n^2 - 7n + 3 =
0

We'll apply the quadratic
formula:

n1 = [7+sqrt(49 -
24)]/4

n1 = (7 + 5)/4

n1 =
3

n2 = (7-5)/4

n2 =
1/2

Since n2 is not a natural number, we'll
accept as solution just n = 3.