[Math] How to know if its permutation or combination in a specific exercise

Question

I have a question,

In how many ways can $6$ tosses of a coin yield $2$ heads and $4$ tails?

Now, to me, the question clearly seems to be of a permutation, as they have asked for the number of ways (they have mentioned nowhere that we have to choose.) But the actual solution tells that the question is solved by a combination.

Why is it a question of combination and not of permutation?

Answer 1

"Permutation" vs. "Combination" is not about "choosing" or "not choosing". It's about whether the order matters or not.

And there is also a difference between how you count and what you are counting.

Here, you are correct: the order "matters" in that you are looking at ordered sequences of six tosses, and you want to count in how many of these ordered sequence there are exactly 2 heads and 4 tails. (If the order of the tosses did not matter, then the answer would be "One: 2 heads and 4 tails").

But, how do you count them? The simplest way of doing this is to imagine you have six blank spots,

    __  __  __  __  __  __

and will place either an H or a T in each spot. You have two Hs and four Ts to put down, and you want to find out in how many ways you can do it. Once you put the two Hs down, the Ts will be forced, so you just need to find in how many ways you can put the two Hs. It doesn't matter which H you put down first. The only thing that matters is where the two Hs end up, not which one went first. So to count this we will use combinations (because order does not matter). And so you need to find the number of ways in which you can choose two of the six spots, which is where the Hs will go. Combinations, because the order in which the Hs are placed does not matter: deciding the first and then the third tosses will be head amounts to the same thing as deciding that the third and the first tosses will be heads.