Solving for an unknown...?

Question

How do I solve C(n,2)=66?

Answer 1

Use the formula for combinations:

C(n,r) = n! / [r!(n-r)!]
So we have: C(n,2) = n!/[2!(n-2)!]
but n! = n*(n-1)*(n-2)!

So use this to simplify C(n,2) as follows:

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

There is an (n-2)! term in the numerator and denominator - they cancel each other out.
We then have:
C(n,2) = n*(n-1) / 2!
= n*(n-1) /2

But we are told that C(n,2) = 66
Therefore, we have:
n*(n-1) /2 = 66
Multiply both sides by 2:
n*(n-1) = 132
Simplify:

n^2 - n = 132
Subtract 132 from both sides:
n^2 - n -132 = 0

Factorize and we have:

(n-12)(n+11) = 0
Therefore: Either (n-12) = 0 OR (n+11)=0
Then either n=12 or n=-11
Since n cannot be negative, we have our answer.
n=12

Answer 2

C(n,2) = n(n-1)/2 = 66, then n(n-1) = 132, so n^2-n-132=0. This factors as (n-12)(n+11) = 0. So n=12

Answer 3

C(n 2) = n! / (2!.(n - r)!) = 66
n! = 132.(n - 2)!
n.(n - 1) = 132
n² - n - 132 = 0
(n - 12).(n + 11) = 0
n = 12 is acceptable answer.