# Combinatorics analysis

A small theater only have $10$ numbered seats. However there is a row of $30$ people to enter into the theater. A group of $3$ friends want to enter into the theater in the same movie session. In how many different ways the $30$ can enter into the theater in order to the $3$ friends could enter in the same session together?

Well, if the theater only have $10$ seats, and there are $30$ people to see the movie. There should be $3$ sessions. As the seats are numbered, in each session the $3$ friends can be sit in $Pr(10,3)$ different ways. But there will be $3$ sessions, they can enter in the $1st$, $2nd$ or in the $3th$ session. So there are $3 \cdot Pr(10,3)$.

But the book's solution is $3! \cdot 10 \cdot 27!$, that I don't understand. Can you help me?Thanks

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We also need to consider the permutations of the other 27 people in the row to enter the theater, so the factor 27! makes sense. But for the other factors I like your solution. – hardmath May 5 '12 at 13:06

## 1 Answer

There are $\binom{10}{3}$ ways to choose the places in the lineup for the $3$ friends so that they can all go to the first showing, and the same number of ways for each of the other two showings. For each of the ways to choose the places, the friends can be placed in $3!$ ways. For each way that this is done, the rest can be placed in $27!$ ways, for a total of $$\binom{10}{3}(3)(3!)(27!).$$

Another way: Call the three friends A, B, and C. There are $30!$ orderings of the people, all equally likely. Whatever position A is in, the probability that B is in the same group of $10$ is $\frac{9}{29}$. Given that A and B are in the same group, the probability that C is also in that group is $\frac{8}{28}$. So the probability all three are in the same group is $\frac{9}{29}\cdot\frac{8}{28}$.

For our count, multiply $\frac{9}{29}\cdot\frac{8}{28}$ by $30!$.

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So you have apllied to a combination, and then a premutation between the $3$ friends. If one simplify the expression you wrote, we'll get $\frac{10!\cdot 3 \cdot 27!}{7!}$. That is the same of $Pr(10,3) \cdot 3\cdot 27!$. So the book's solution is wrong.Thanks for the help – João May 5 '12 at 15:35
@João: I cannot think of an interpretation that would make the book's answer correct. Your analysis was correct, apart from the minor oversight about $27!$. The answer I wrote is a minor variant of that analysis. – André Nicolas May 5 '12 at 15:48