What is n if the number of combinations of (n+1) distinct elements taken 2 at a time is 66.

Let the number of combinations of (n+1) distinct elements
taken 2 at a time be C(n+1 , 2).

We know, from enunciation,
that C(n+1 , 2) = 66

We'll recall how to write the number
of combinations of n elements taken k at a time in terms of
factorial:

C(n,k) =
n!/k!(n-k)!

According to this formula, we'll
have:

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

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

But (n+1)! =
(n-1)!*n*(n+1)

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

We'll simplify and we'll
get:

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

But C(n+1 , 2) =
66;

n*(n+1)/2! = 66

n*(n+1) =
1*2*66

We'll remove the
brackets:

n^2 + n - 132 =
0

We'll apply quadratic
formula:

n1 =
[-1+sqrt(1+528)]/2

n1 =
(-1+23)/2

n1 = 11

n2 =
-12

Since n has to be a natural number, we'll
reject the negative value of n and we'll keep n =
11.