Why do we use combinations instead of permutation when finding amount of ways of flipping coins 
As on the bottom of this question there is a link in which it asked for the number of combinations with exactly three heads and three heads or more, why does the answer utilize combinations?

When looking at the math problem, I was confused due to the fact that I thought that the order in flipping the coins mattered but the answer didn't make sense to me very much because HHHTTTTT is different from HTHHTTTT
But according to numerous sources, the combination of flipping exactly three heads in a row is 8C3 which I didn't understand, because I thought that I was going to be using permutations because order matters? 
I believed that finding the number of flipping exactly three heads was going to be 8P3?
(Link to the Math Problem I am Referring to)
 A: Take a simpler case that we can do by hand.  Say we want two heads out of four tosses.  The combination approach says there should be ${4 \choose 2}=6$ ways to get that, which are $HHTT, HTHT, HTTH, THHT, THTH, TTHH$.  If you use permutations there should be $4 \cdot 3=12$ ways, but there are not.  The point is that the two heads are equivalent, so choosing that the first flip should be heads and then the third should be heads gives $HTHT$.  If you choose the third flip should be heads and then the first should be heads, you again get $HTHT$ so you double count with permutations.
A: The order does matter. The reasoning behind the answer being $\binom{8}{3}$ $(=$ 8C3$)$ is as follows: a sequence of $8$ coin flips in which exactly $3$ of the flips are heads is uniquely determined by the choice of $3$ of the $8$ flips; the chosen flips are heads, and the rest are tails.
How does this differ from 8P3? By using 8P3, you implicitly state that the order of the choice of flips also matters (not just the order of the flips themselves). So for instance, in this scenario, choosing flips number $1,2$ and $3$ to be heads would not be the same as choosing flips number $3,2$ and $1$.
For any three distinct flips, there are $3!$ distinct ways to order them. Combinations compensate this overcounting precisely by dividing by $3!$. It considers the two choices above as the same.
