[Math] Combinations confusion for coin flips

combinationspermutations

10 fair coins are tossed. How many outcomes have 3 Heads?
I'm supposed to solve it with combination C(10, 3). But…

How do you know it's a combination that will solve it? I'm not interested in what makes it a combination, instead of a permutation. I know the answer is (some #)/3^10 total outcomes. But what is your thought process that initially makes you think, "I need to use the (n!)/(k!(n-k)!) combination formula on it."? I can easily identify when to use combinations on every combination-required problem I've encountered except for coin tosses.

I've already looked at Ian's problem, but our confusion seems a little different:
Combinations and Permutations in coin tossing

I have no problems understanding any other permutation or combination problems, like the (common?) horse race ordering problem, or picking colored balls out of urns. But something about coin flip combinations just completely baffles me. It might have something to do with the 50/50 heads tails chance.

Best Answer

The $50:50$ odds on each toss means that all distinct outcomes are equally weighted, and so we don't worry about it any further. Here an outcome is an arrangement of heads and tails in ten coin tosses. Hence we are counting permutations.

So to count the favoured outcomes, which are arrangements that contain exactly $3$ heads, consider that the task is to select $3$ of $10$ 'places' to put the heads, and put tails in the remainder. That's $10$ choose $3$ ways: $^{10}\mathrm C_3$.

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

"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.

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

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