Heads or Tails?

We have 20 identical coins. The number of ways to place 14 coins heads up and 6 coins tails up in a row is:

${20 \choose 14} = 38760$

The restriction in the number of positions is that no two coins that are tails up are right next to one another. First place the 6 coins that are tails up in a row which can be done in only one way because each coin is indistinguishable.

Between each two of these 6 coins we must place a coin which is heads up. So we must place 5 coins heads up between the 6 coins that are tails up. One coin must be placed in each of the positions indicated by $A, B, C, D, E$ .

The restriction is now met and the result is shown below.

Now each of the remaining 14 - 5 = 9 coins that are heads up can be placed in one of the indicated 7 spots:

$A, B, C, D, E \quad \text{or} \quad X_1 \quad \text{or} \quad X_2$

The number of ways this can be done is:

${9 + 7 -1 \choose 9} = {15 \choose 9} = 5005$

Finally the probability that this happens, rounded of to the nearest thousandth is:

$\frac{{15 \choose 9}}{{20 \choose 14}} = \frac{5005}{38760} \approx 0.129$