[Math] The number of ways a coin can be tossed 6 times so that there is exactly 3 heads and no two heads occur in a row is

probability

I am trying to figure out the problem below.

The number of ways a coin can be tossed 6 times so that there is exactly 3 heads and no two heads occur in a row is?

At first it seemed like it would just be .5•.5•.5•.5•.5•.5 which would be 0.015625, since none of the outcomes are dependent on each other. Then as I was reading the question more carefully, I thought maybe the answer is 2? Since there are only 2 outcomes for each toss?

All help greatly appreciated.

Best Answer

The question asks for "The number of ways" that you can flip six coins with exactly 3 heads and 3 tails such that no two heads appear next to one another.

Use the tails as a barrier:

$$\underline{~}T\underline{~}T\underline{~}T\underline{~}$$

The three heads may go into those four available spaces with at most one head per space. Choose which three spaces are occupied.

Four spaces, you want to choose three of them, there are then $\binom{4}{3}=4$ different such sequences. The sequences are specifically HTHTHT, HTHTTH, HTTHTH, THTHTH

As an aside, the probability of this happening is: The number of ways it could happen / The total number of ways of flipping coins regardless of this happening.

$\frac{4}{2^6} = \frac{1}{2^4} = \frac{1}{16}$

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