Bernoulli's coin flips

This is what is given $\Rightarrow$ _ ? TH ? TH ? TH ? TH ? TH ? _ , Every _ '?' _ can be filled with one or more _ 'H' 's or 'T' 's _ .

We have now 3 H _ and _ 2 T , If we put _ one or more H _ in left most _ '?' _ , we can not achieve target of _ 3 HH _ .

Also , if we put one or more _ one or more T _ in right most _'?' _ , we can not achieve goal .

Hence, we have five places for 3 H _ and _ five places for 2 T .

We can place all _ T's _ together or can place them separate .

There are $5 + \binom{5}{2} = 15$ ways to do this .

We can place all _ H's _ together or can place as _{1 H , 2H} _ or all separate .

There are $5 + 2\binom {5}{2} + \binom {5}{3} = 35$ ways to do this .

Hence, there are _ 15 *35 = 525 _ sequences that could be written down .

He has $5$ instances of consecutive $TH, 4$ instances of consecutive $HT, 3$ instances of consecutive $HH$ , and $2$ instances of consecutive $TT$ . Multiplying this out we get (with overcount) $15$ $H's$ and $13$ $T's$ .

Since both these numbers are odd we know that one of each must be at the endpoints meaning we have $8$ $H's$ and $7$ $T's$ . Playing around with the possibilities we get that the sequence has to start with a $T$ and end with a $H$ .

Now lets write the sequence with just the $H's$ and a $T$ at the beginning so it looks like this $THHHHHHHH$

Lets deal with the $5$ instances of consecutive $TH, 4$ instances of consecutive $HT$ first

We can fill in $4$ $T's$ in any of the $7$ spaces between consecutive $H's$ to get $\binom{7}{4} = 35$ which would satisfy all of the conditions except the $2$ $TT$

Now we have two cases

$\bullet$ Case 1 : $2$ $TT$ in different spots

For this case we choose the two locations after a $T$ to place another $T$ which means there are $\binom{5}{2} = 10$ options here.

$\bullet$ Case 2 : $TTT$

For this case we have to choose the location after a $T$ to place a $TT$ which adds $\binom{5}{1} = 5$ extra options.

Therefore we get $35 \times (10 + 5) = \boxed{525}$

Consider the 10-flip sequence THTHTHTHTH. There are 5 instances of TH and 4 instances of HT, which is what we want in our 15-flip sequences. To create a 15-flip sequence with 3 instances of HH and 2 instances of TT, we just add 3 H's next to another H and 2 T's next to another T.

There are 5 places where there is an H and 3 new H's to put in, so there are $\binom{5+3-1}{3} = \binom{7}{3}$ ways to place the 3 H's. Similarly, there are 5 places where there is a T and 2 new T's to put in, so there are $\binom{5+2-1}{2} = \binom{6}{2}$ ways to place the T's. In total, there are $\binom{7}{3} \binom{6}{2} = 35*15 = \boxed{525}$ ways to create the sequences.