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

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### 2 Answers

The number of combinations of two elements taken from a set containing n elements is given by C(n, 2) = n!/2!*(n - 2)!

It is given that C(n + 1, 2) = (n + 1)!/2!*(n - 1)! = 66

=> (n + 1)n/2 = 66

=> n^2 + n = 132

=> n^2 + n - 132 = 0

=> n^2 + 12n - 11n - 132 = 0

=> n(n + 12) - 11(n + 12) = 0

=> (n - 11)(n + 12) = 0

=> n = 11 and n = -12

As the number of elements cannot be negative take only n = 11

**The value of n = 11.**

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.**