How many people are there? 66 hand shakes on a party. If each person shakes every body's hand, how many people were at that party? (I have the answer as 12 people but through use of elementary logic) Could any one suggest a more mathematical way of handling this?
 A: The number of handshakes is n(n-1)/2 where n is the number of people. Each of the n people make n-1 handshakes, so the answer would be n(n-1) but we divide by 2 because we double counted each handshake: we thought that A shaking hands with B is different from the reverse case, but they're not. 
A: The number of possible handshakes is equal to $\dbinom{n}{2}$. These are the combinations of $n$ over $2$. (With combinations you count the ways of selecting $2$ persons out of $n$ so that order does not matter.) Thus you need to solve the equation $$66=\dbinom{n}{2} \iff 66=\frac{n!}{2!(n-2)!} \iff 66=\frac{n(n-1)(n-2)!}{\phantom{(n-1)}2(n-2)!} \iff 132=n(n-1)$$ which is equivalent to the quadratic equation $$n^2-n-132=0 \iff (n+11)(n-12)=0$$ with roots $n=-11$ (rejected, as $n>0$) and $n=12$ which is accepted.
A: With n amount of people there are $\frac{1}{2}(n^2 +n)$ amount of handshakes because the amount of handshakes is a triangle number, with base $n$. solving that however you may (quadratic formula, fractionation CTS) gives you solutions $132=n^2+n$,  $n^2+n-1380=0$ factorizes to $(n-12)(n+11)$ solution are $+12$ or $-11$. There is no such thing as a negative amount of people so we ignore that solution. 
