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I tried the most ubiquitous taught way. Bundle two heads together {HH}. Now, we have H,{HH},T,T. We have $\frac{4!}{2!}$ combinations if we consider H and {HH} distinct. So in the previous calculation H{HH}TT and {HH}HTT are both counted (and considered different). Basically 3 consecutive heads are counted twice. To arrive at the correct answer, I subtract the cases when 3 heads are consecutive. The final answer will be

$ \frac{4!}{2!}-\frac{3!}{2!} = 9 $

I wanted to ask if there is a general way to solve such problems. For example, the problem would be much difficult if we had 10 heads and 5 tails, and we want to enumerate the items in which there are at least 3 consecutive heads.

SSB
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  • The problem is not clear. Are you just saying, "find the number of arrangements in which there are at least two consecutive Heads"? If so...well, the only bad arrangement is $HTHTH$. – lulu Apr 29 '21 at 14:39
  • I think it is clarified in the statement and the solution is given. The purpose is more geared towards a generalized approach of solving such issues. – SSB Apr 29 '21 at 14:52
  • I couldn't follow your solution, it seems unnecessarily complicated. For a general situation, consider the lists of the form $T^{a_1}HT^{a_2}H\cdots HT^{a_{N+1}}$ (assuming you have $N$ Heads). Where $\sum a_i= M$ (if you have $M$ Tails) and $a_2, \cdots, a_{N}$ are all strictly positive. Counting these is a routine Stars and Bars problem. Then just subtract from the unrestricted arrangements. – lulu Apr 29 '21 at 15:01

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