Question 647499
I assume you mean (9 choose 3). By definition c(n,r) or (n choose r) is  {{{(n!)/((n-r)!*r!)}}}

This sounds intimidating I know, but if you think about it, it isn't terrible.

Recall that a permutation is {{{n!/(n-r)!}}}, because for instance if we want to figure out the number of ways between 9 horses there can be a trifecta, then you would start with 9 horses to choose from, then 8, and then 7. Another way to think about this is you have {{{9!/(9-3)! =  9!/6! = 9*8*7}}}since you are eliminating the other 6 horses each time.  Order matters in this case. For combinations, order doesn't matter. So, we factor out the order. Let's say you win if any of the three horses you pick takes win,place,and show. Well if you win, between horse A,B,C you could have win,place,show; w,s,p; s,w,p ... etc.  There are 3! ways to do this. So let's divide it out.  So we get  {{{9!/(9-3)!3!}}} =  C(9,3).

So this is  {{{(9*8*7*6*5*4*3*2*1)/((6*5*4*3*2*1)*(3*2*1))}}} = {{{(9*8*7*cross(6)*cross(5)*cross(4)*cross(3)*cross(2)*cross(1))/(cross(6)*cross(5)*cross(4)*cross(3)*cross(2)*cross(1)*3*2*1)}}} = {{{(9*8*7)/6}}} = {{{(3*8*7)/2}}} = {{{3*4*7}}} = 84.

C(9,3) = 84