# Seating children in the cinema

I just had finished my class and have been struggling with a problem.

There's $9$ seats in the cinema, and two families $F_a=\{F_1,F_2,F_3,F_4,F_5\},$ $F_b=\{F_a,F_b,F_c,F_d\}$

In how many ways can the kids be sitting together?

The answer is $5!\times5!$ apparently, but I can't figure out how you get to that solution. There's obviously $5!$ ways to order them, but you can also put the kids in 5 different positions (since there's 9 seats available). Shouldn't the answer already start differing?

$5\times5!$, but that's without considering the order of the rest of the people (the 4 parents). They can be arranged in $4!$ ways, without caring about the order of the kids. So now I'm at $5\times4!\times5!$, which I'm guessing is not the correct answer.

• You're almost there. Note that 5*4! = 5! Commented May 19, 2016 at 20:02
• I agree. 5 positions for the leftmost kid, $5!$ ways of ordering the kids, $4!$ ways of ordering the parents. Total $5!5!$. Commented May 19, 2016 at 20:03
• I just had the realization and was about to delete my silly question, but I'll leave it to remind me of my own stupidity. Commented May 19, 2016 at 20:51

You have the correct analysis. Here is another way to get $5!5!$.
We will assume these are families in the old-style definition, a mommy and a daddy and kids. We have the $4$ adults, and the block B of kids, $5$ objects altogether. These can be arranged in $5!$ ways. But for each of these ways, the kids can be arranged within the block in $5!$ ways, for a total of $5!5!$.
If you have two parents in each family and are asking how many ways the five kids can sit together, you have $5$ choices for the seat of the leftmost kid, $5!$ ways to order the kids in the five seats starting from the leftmost, and $4!$ ways to order the parents in the remaining seats, giving $5 \cdot 5! \cdot 4!=5! \cdot 5!$