Combinations (Hand Shakes) I have a question as such:

You and a group of 9 Friends are playing a game of basketball. At the end of the game, if each player shakes hands with every other player, how many handshakes will there be?

At first I interpreted this question logically, 
if you have 9 friends, you yourself will shake hands with 9 other people (1 less because you don't include yourself). 

The other 9 friends will shake hands with 9 other people.
Thus being 9*9 = 81 handshakes.

But the answer is 45

( i.e $10\choose2$) 

which logically to me, makes absolutely no sense(if every person if shaking everyone else's hand)
Any clarification would be much appreciated.
There is one handshake for each pair of players, and there are $\binom{10}2$ pairs of players. Your calculation of of $9\cdot9$ is off in two respects. First, there are $10$ players, not $9$, and each shakes hands $9$ times, so your reasoning should result in $10\cdot9=90$ handshakes, not $9^2=81$. But you’ve counted each handshake twice, once for each of the two players involved in it, so you have to divide that $90$ by $2$. And when you do that, you do indeed get $45=\binom{10}2$.
Answer is 36. Because on each hand shakes we tempt to include ourselves, but we should not be doing that. Instead we should calculate one less hand shakes. Therefore re by the equation n*(n-1)/2 we could find the solution. That is n=9 and therefore the answer is 36.