# Fair coin probability question

A fair coin is tossed $20$ times. Find the probability that exactly $6$ heads were tossed one of which was tossed on the first toss and one of which was tossed on the last toss, and that no consecutive tosses were heads.

Typically, the probability would be ${20 \choose 6}(1/2)^{6}(1/2)^{14}$.

However, we are restricted in that we cannot include all combinations of successes and failures.

${10 \choose 4}(0.5)^4 (1.5)^{14}$?

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Just count how many of the $2^{20}$ possible sequences satisfy the conditions. Every legal sequence must have the form HT...H, such that the "..." consists of 4 "HT" combinations and 9 "T"s that are not part of such a combination.