Suppose a coin is tossed $12$ times and there are $3$ heads and $9$ tails. How many sequences are there in which there are at least $5$ tails in a row?
I know this is Permutation with repetition. My line of thinking is we start at $5$ tails in a row, followed by $7$ flips being permuted. $TTTTTXXXXXXX$ or $P(7)$
Then we take the next case, $6$ tails in a row, followed by $6$ flips being permuted. $TTTTTTXXXXXX$ or $P(6)$
We continue this pattern until we get to $9$ $T$ in a row, followed by $3$ flips being permuted, where the formula ends up being:$$P(7)+P(6)+P(5)+P(4)+P(3)$$$$=7!+6!+5!+4!+3!$$However, we have to account for double counting, since switching a $T$ with another $T$ or an $H$ with another $H$ doesn't change the sequence. So we have:$$\frac{7!+6!+5!+4!+3!}{2}=2,955$$I feel like my logic on this is pretty sound, however my teacher's answer key for the practice final says the answer should be $$4\cdot {7\choose 4}=140$$Where am I going wrong here?
Best Answer
If we line up the 3 heads, this creates 4 gaps into which we can put the tails.
1) There are 4 ways to select the gap which will contain at least 5 tails, so put 5 tails in this gap.
2) There are $\dbinom{7}{3}$ ways to distribute the remaining 4 tails, since there are 4 tails and 3 dividers (the heads).
This gives $4\dbinom{7}{3}$ possibilities.