A car can hold 3 people in the front seat and 4 in the back seat. In how many ways can 7 people be seated in the car if 2 particular people must sit in the back seat and 1 particular person is the driver?

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- Jul 12th 2008, 12:23 PMMisa-CampoPermutations again
A car can hold 3 people in the front seat and 4 in the back seat. In how many ways can 7 people be seated in the car if 2 particular people must sit in the back seat and 1 particular person is the driver?

- Jul 12th 2008, 01:32 PMPlato
- Jul 12th 2008, 01:43 PMparticlejohn
Shouldn't it be $\displaystyle 4 \binom{3}{2} \binom{4}{2} = 72$? Because you fix the $\displaystyle 2 $ people in the back seat and the $\displaystyle 1 $ person in the front seat?

- Jul 12th 2008, 02:25 PMjanvdl
If you are getting help, don't scoff at other members or be rude. Thanks.

- Jul 12th 2008, 02:26 PMPlato
The driver is fixed. That means there are six people left.

Since two of the six insist on sitting in the back, we have only four people from which can choose the other two for the front seat. $\displaystyle \binom {4} {2}$ and two ways to arrange those two. At this point we have four people remaining for the back seat, $\displaystyle 4!$. - Jul 12th 2008, 02:31 PMparticlejohn

Or could you fix the $\displaystyle 2 $ particular people in the back seat? That means there are $\displaystyle 5 $ people left. $\displaystyle 1 $ must be the driver. So we have $\displaystyle 4 $ people from which we can choose the other $\displaystyle 2 $ for the front seat. Same answer: $\displaystyle 2\cdot 4! \binom{4}{2} $. - Jul 12th 2008, 02:36 PMMisa-Campo:d
well sorry but i had no intention, was just being straightforward thats all, ill be more polite next time lol

- Jul 12th 2008, 02:39 PMPlato