# Counting Question

I'm taking a crack at learning probability and statistics starting from the basics. Anyways here is the question.

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A fair coin is tossed 10 times and the sequence of scores recorded.

How many sequences are there?

How many sequences are there that contain exactly 3 heads?

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The first one is no problem. I understand it. It simply is 2^10 = 1024 possible sequences.

The second one is the one i have trouble dealing with. How do you compute it with that special case added in? The answer given is 120.

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eg THHHTTTT.... THTT...THT..TTH. your counting the exact number of sequeences that have only ("exactly") 3 HHH. not necessary in successive order(that is THHHT)

mathwurkz said:
A fair coin is tossed 10 times and the sequence of scores recorded.

How many sequences are there?

How many sequences are there that contain exactly 3 heads?
For the second one, you have exactly 3 heads.

_, _, _, _, _, _, _, _, _, _

Each spot corresponds to a coin toss. First spot corresponds to the first toss, the second spot the second toss, etc.

You have 3 of the ten spots you need to put a head. One way to do this is to put a head in the first, second and third spots. Another way to do it is to place it one in the first spot one in the third spot and one in the last spot. You could list all the 120 possible ways to put 3 heads in 10 spots but there is an easier way.

If you have looked at combinations at all then by now you should see that the answer is just 10 choose 3.

$$\frac{10!}{3!7!} = 120$$

If you have not looked at combinations then I would recommend you study combinations before trying to work this kind of problem out.

Best of luck,

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