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Suppose a fair coin which has 2 faces-Head(H) and Tail(T) is thrown 12 times and we get exactly 5 heads: $$H,H,H,H,H,T,T,T,T,T,T,T$$ How can I find the number of different arrangements it can have? One possible arrangement:

$$H,T,T,H,T,T,H,H,T,T,T,H$$ Another possible arrangement: $$ H,H,H,T,T,T,T,T,T,T,H,H$$

I can find the probability of getting exactly 5 heads by using binomial distribution but i cannot find the actual number of arrangements.

Irtiza
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  • Please negative voter,tell the reason. – Irtiza Apr 29 '15 at 14:06
  • I am not the downvoter, but I agree that this question has issues. One, You start to talk about a dice, and the title is about a coin. Two, you did not show your work so far: what did you try, how far did you get, where are you stuck? – 5xum Apr 29 '15 at 14:10
  • Another problem is that this just looks like a homework question. (I am not the negative voter by the way. I have seen a lot posts which are worse) – Jack Yoon Apr 29 '15 at 14:11
  • i hope this is fine now – Irtiza Apr 29 '15 at 14:20

1 Answers1

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It is the binomial coefficient $\binom{12}{5}$, i.e., $\dfrac{12!}{5!(12-5)!}$. This corresponds to the number of different ways of placing the $5$ heads in the $12$ positions.

JMP
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inwit
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