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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.

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Kawaii
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    We are counting how many words of length $10$ there are over the alphabet {H,T}, that have exactly $3$ occurrences of H. To count how many, choose the $3$ places in the string that the H's will occupy. By definition this can be done in $\binom{10}{3}$ ways. There are $2^{10}$ possible strings of H and T, all equally likely. So our probability is $\frac{\binom{10}{3}}{2^{10}}$. – André Nicolas Oct 07 '15 at 01:07
  • But if instead, I was counting how many words have 3 occurences of T, wouldn't it still be C(10, 3)? How do you differentiate? – Kawaii Oct 07 '15 at 01:35
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    You don't. The probability of $3$ heads out of $10$ is the same as the probability of $3$ tails out of $10$, if the coin is fair. – André Nicolas Oct 07 '15 at 02:11
  • I now see my problem. In another example choose 2 heads for 3 tosses, I was confused with the 2 in 2^3 total outcomes, and the 2 from choosing 2 heads. – Kawaii Oct 07 '15 at 17:04

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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: $^{10}\mathrm C_3$.

Graham Kemp
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think about naming the order of the tosses ... toss#1, toss#2 etc.

e.g. The number of ways of getting 3 heads when tossing 5 coins is the same as the number of ways of deciding which 3 of the 5 tosses came up heads

e.g. the choice ${2,4,5}$ corresponds to the sequence THTHH

the choice ${1,2,5}$ corresponds to the sequence HHTTH

so the number of sequences containing 3 heads =$\binom 52$

WW1
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  • There are $10$ coins tossed. If this is intended as a simpler example, that should be stated somewhere. – Marconius Oct 07 '15 at 02:02