Combinations/Permutations intuition question I have a relatively question that I am struggling with that is an intuition question and I'm hoping someone can help. 
I start with the basic and simple question, how many ways can you get 3 heads EXACTLY with 5 flips. The math behind this simple: You have 5 slots for the first heads, 4 for the second, 3 for the 3rd = 5*4*3= 60. However, we have to divide this by 6 because order doesn't matter. While I somewhat understand I am struggling with the intuition so my question is. 
What does the 60 represent? 60 represents if the order matters correct, but what larger set is 60 a subset of? Like in simple english what set is a 60 a subset of? 
 A: In this situation, $60$ is the number of ways to choose which flips will be heads, where we count "first flip, third flip, fourth flip" as different from "third flip, fourth flip, first flip". If the three heads were somehow distinct, then $60$ would be the number of ways in which they could occur.
More concretely, $60$ is the number of possible arrangements of the symbols, $H_1, H_2, H_3, T, T$. For a bigger-picture perspective, $5!=120$ is the number of arrangements of the symbols, $H_1, H_2, H_3, T_1, T_2$. We divide by $2!$ because we don't care which tail is which, and we divide by $3!$ because we don't care which head is which. Thus: $\binom53=\frac{5!}{3!2!}$.
A: To make the picture simpler, take exactly TWO heads in $5$ flips.
Then you can place
 - 1st H in position $1$, and the 2nd in one of the remaining $4$
 - 1st H in position $2$, and the 2nd in one of the remaining $3$
 -  ...
that is in
$4+3+2+1=10=5/2*4={5 \choose 2}=5*4/2$
ways.
If the problem was in how many ways you can place $1$ black, $1$ red and $3$ white marbles in a row, then this is $5*4=20$.
Now, if the red marble becomes black, you have to divide the above by $2=2!$,
and you get a situation which is analogous to the two heads above.
