My question simply is how many possible different combinations of any $18$ numbers can be made from $136$ numbers?
Number of different combinations of $18$ numbers is $18! = 6402373705728000$
I do not need that, but once again number of different $18$ numbers that can be made out of any $136$ numbers.
There are $136$ choice for the first choice and $135$ for the second and so on so there are $136*135*....*119=\frac {136!}{118!}$ ways to select 18 items.
But we don't care about order. If we pick 18 items it does not matter what order we seected them in. And for any 18 items there are $18! $ orders we could have picked them.
The above number includes all the ways in every possible order. $18! $ times more than we want.
So we want $\dfrac {\frac {136!}{118!}}{18!}=\frac {136!}{18!118!} $.
In general, to select $a$ from $n $ options without regard to order, there are $\frac {n!}{a! (n-a)!} $ ways to do this. We write that as ${n\choose a}$ and we call it "$n $ choose $a $".
