Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So I have this question regarding counting the possible outcomes of a coin flipping sequence. Here it is:

A coin is flipped 12 times where each flip comes up either as heads or tails. How many possible outcomes contain exactly 3 heads, where each head is immediately followed by at least 2 tails?

So I figure the outcome string must contain a combination of the following string blocks:


For example you could use HTT + HTTT + HTTT, which gives you three heads with each of them having at least two proceeding tails. The length of this string is 11, meaning there is one more possible outcome that has to be a tail. This can be placed at the beginning of the string, end of the string, or in between each string block. Thus the possible outcome strings are:


How do I systematically count all the possible outcomes strings using a combination of the string blocks HTT, and HTTT? How do I avoid counting possible repeats?

share|cite|improve this question

Since each head is followed by two tails, you can make "HTT" a block and regard it as one single element. Now, you have three "HTT" blocks and three "T" left. So the way of arrangement should be $6$ choose $3$, which is $20$.

share|cite|improve this answer

We have three heads H, followed by at least two tails T, which means we have three blocks HTT. We have no more heads, so the rest must be tails as well. There are 4 spots where we can put those Ts:

x HTT x HTT x HTT x

At every x there can be 0 or more Ts, for a total of 3 Ts. That can be done in three ways:

  • 3 Ts at one spot, for a total of $4$ spots
  • 2 Ts at one spot, 1 T at another spot, for a total of $4 \times 3$ spots
  • 1 T at three different spots, with one spot left empty, for a total of $4$ spots that can be left empty

This totals to $4 + 4 \times 3 + 4 = 20$ different combinations.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.