Answer: 560

Small Hint:

Compare the numbers of $\mathrm{HT}$ and $\mathrm{TH}$ transitions to determine the first and last tosses

Big Hint:

Translate the $\mathrm{HH}$ and $\mathrm{TT}$ counts into totals distributed among alternating runs

Solution:

Since there are four $\mathrm{TH}$ transitions and three $\mathrm{HT}$ transitions, every valid sequence starts with $\mathrm{T}$ and ends with $\mathrm{H}.$ It therefore has four $\mathrm{T}$-runs and four $\mathrm{H}$-runs, alternating.

If the $\mathrm{H}$-runs have total length $h,$ then the number of $\mathrm{HH}$ transitions is $h-4.$ Thus $h=6,$ and the positive lengths of the four $\mathrm{H}$-runs can be chosen in $\binom{6-1}{4-1}=\binom53=10$ ways. Similarly, five $\mathrm{TT}$ transitions mean that the four $\mathrm{T}$-runs have total length $9,$ giving $\binom{9-1}{4-1}=\binom83=56$ choices. The alternating order is fixed, so the number of sequences is $10\cdot56=560.$