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A coin is flipped eight times where each flip comes up either heads or tails. How many possible outcomes

a) are there in total?

b) contain exactly three heads?

c) contain at least three heads?

d) contain the same number of heads and tails?

the order does matter here!

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I’ve changed the title: the question isn’t about permutations. –  Brian M. Scott Nov 3 '12 at 19:56
3  
Since this is homework, have you tried anything that you could show us? –  Jean-Sébastien Nov 3 '12 at 19:56
    
the order does matter here! –  Alpha Nov 3 '12 at 19:56
    
I have no idea how to solve... –  Alpha Nov 3 '12 at 19:58
7  
Do you know about accepting answers? All of the questions can be answered by brute force if you can’t think of anything better, but surely you can at least answer (a); where are you stuck on the others? –  Brian M. Scott Nov 3 '12 at 19:58

1 Answer 1

I'm doing $b)$ for you, in two different, yet similar way. You want $3$ heads, which implicitly means you will have $5$ tails. We can see this as permutations of the words $HHHTTTTT$. Since we have similar objects, this is done in $$ \frac{8!}{3!5!} $$ ways. I believe this is where you see permutations

Here is another method that yields the same result. You have $3$ heads to place into $8$ slots, the remaining $5$ must be tails. The number of ways to choose where the heads go is given by $$ {8\choose 3}=\frac{8!}{3!5!}. $$ Note that we could have chosen where we want to place the tails, in ${8\choose 5}$ ways, which gives the same thing. This is where Brian says it is not really permutations, but combinations.

Can you figure out the rest now?

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