# Permutations of the word $ENTERTAINMENT$ with restrictions?

### Hi, the question states:

The letters of $ENTERTAINMENT$ are arranged in a row. Find the probability that two $E$ 's are together, and one is apart.

The answers is $5/13$, but I keep obtaining $45/143$.

My process of the question is as follows:

1) I created a group for the restrictions that looked like: $_EE_$ where each underscore represents a letter other than that third $E$ (due to the restriction that only two $E$ 's can be next to each other. This group was ordered as $3P2 * 10P2$.

2) I then ordered all the groups, and my final answer for all the arrangements (including repeated letters) was: $(10!*3P2*10P2)/(3!*3!*3!)$

3) I then divided this by the sample space since it is a probability question (also taking into account the repitition), and got $45/153$, which is the wrong answer.

What am I doing wrong? Any help would be greatly appreciated.

• How can you guarantee that the other one E is apart in your solution? Commented Oct 26, 2014 at 14:22
• Try see $_EE_$ as a "unique letter", then compute the number of possible permutations for this "letter". Commented Oct 26, 2014 at 14:23
• The only way I could think of, was the way I stated; i.e. enclose two E's with two other letters. Is there another way? This seems to be a recurring problem I cannot resolve, so if you are able to tell me, I would be highly thankful. Commented Oct 26, 2014 at 14:23
• @DiegoMath I did that, but which step in my calculations did I do something incorrect? Commented Oct 26, 2014 at 14:24
• I may have done my repetitions incorrectly. I'll retry the question Commented Oct 26, 2014 at 14:26

To begin, there are $\binom{13}{3}$ different choices of slots for the $3$ $E$'s to be in. If two of the $E$'s must be adjacent, then we can think of them as forming a single composite letter $EE$ and shrink the total number of letters from $13$ to $12$. In this manner, there are $10$ slots which $EE$ can be in such that it is not at either end, as well as $2$ end slots. If $EE$ is in any of the $10$ non-end slots, the last $E$ can be in any of $12-1-2=9$ slots such that $E$ is not adjacent to $EE$. If $EE$ is at either end slot, there are $12-1-1=10$ slots such that $E$ is not adjacent to $EE$.
It follows that the probability that two $E$'s are adjacent but the last $E$ is not adjacent to either of the other two $E$'s is $\frac{10\cdot 9 + 2\cdot 10}{\binom{13}{3}}=\frac{5}{13}$.
Take all the 10 letters excluding the Es and they can be arranged in $10!/\{3!*3!\}$ ways. Glue two Es together and insert this and the other E into the 11 spaces between the letters (including beginning and end) in $11P2$ ways. Then divide by the total number of free permutations $13!/{3!}^3$ and you get 5/13.