# During a party a total of 78 handshakes occurred. If each person shook hands once with each of the other people , how many people were at the party?

##### 1 Answer
Jun 21, 2018

$n = 13$

#### Explanation:

Lets say we have $n$ people. The first one can shake hands with $n - 1$ people (everyone, except himself). The next person can shake hands with $n - 2$ (everyone, except person 1 and himself). The one after that with $n - 3$ and so on. So the sum would be:
$1 + 2 + 3 + 4. . . + \left(n - 2\right) + \left(n - 1\right) = \text{total sum of handshakes}$
This can be expressed by:
$\frac{\left(n - 1\right) n}{2} = \text{total sum of handshakes}$
$\frac{\left(n - 1\right) n}{2} = 78 | \cdot 2$
$\left(n - 1\right) n = 156$
$n = 13$