# Suppose a coin in tossed $12$ times and there are $3$ heads and $9$ tails. How many sequences…

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?

• Does dividing by $2$ eliminate all the repetitions? Switching a $T$ with another $T$ or an $H$ with another $H$ doesn't change the sequence. But there are many ways to switch, so much more repetitions are added. – Colescu May 6 '16 at 5:28
• Your $2955$ is too large. There are only $\binom{12}{3}=220$ head/tail sequences with $3$ heads and $9$ tails. – André Nicolas May 6 '16 at 5:29
• How would it be ${12 \choose 3} = 220$? That would mean out of the $12$ flips, choose a group of $3$ flips. Since it's permutation with repetition it would be ${12\choose 9, 3}={12\choose 9}{3\choose 3}={12\choose 9}\cdot 1$ – Jodo1992 May 6 '16 at 5:49
• Oh okay, I just realized that ${12\choose 9}$ and ${12\choose 3}$ are equivalent -_-. Still, how do I take the problem from that point? – Jodo1992 May 6 '16 at 5:50

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.

• I think this finally solidified what @Woria was trying to describe to me. Thank you – Jodo1992 May 6 '16 at 17:05

Hint: You can construct such sequence in the following consecutive steps:

Step 1: construct a sequence of $3$ Heads and $4$ Tails;

Step 2: Now put a block of extra $5$ Tails in the sequence.

• So Step 1 would yield ${7\choose 4,3}={7\choose 4}\cdot 1$, but isn't Step 2 unnecessary since we do not need to compute the permutation of similar Tails? – Jodo1992 May 6 '16 at 6:33
• Oh I see it now. So then I would continue with ${6\choose 3,3}={6\choose 3}\cdot 1$, then ${5\choose 2,3}={5\choose 3}\cdot 1$ etc, yes? – Jodo1992 May 6 '16 at 6:35
• For that block of Step 2 you have $4$ positions: $*H*H*H*$. – Woria May 6 '16 at 6:42
• I'm sorry, I'm not grasping what you mean by that. Aren't the extra $5$ tails all together? – Jodo1992 May 6 '16 at 6:44
• Let's start with a sequence from Step 1, for example: $$TTETEET$$ Now in Step 2 by using that block $B$ we can construct $4$ sequences: $$BTTETEET, TTEBTEET, TTETEBET, TTETEEBT$$ – Woria May 6 '16 at 6:47