# Permutations again

• Jul 12th 2008, 12:23 PM
Misa-Campo
Permutations 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 PM
Plato
Quote:

Originally Posted by Misa-Campo
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?

(You changed the question on me!)

There are $\displaystyle \binom {4} {2} = 6$ ways to select the two other persons in the front seat.
They can be arranged in 2 ways. The other four in the back can be arranged in (4!).
• Jul 12th 2008, 01:43 PM
particlejohn
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 PM
janvdl
If you are getting help, don't scoff at other members or be rude. Thanks.
• Jul 12th 2008, 02:26 PM
Plato
Quote:

Originally Posted by particlejohn
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?

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 PM
particlejohn
Quote:

Originally Posted by Plato
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, $\displaystyle 4!$.

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 PM
Misa-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 PM
Plato
Quote:

Originally Posted by particlejohn
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: $\displaystyle 2\cdot 4! \binom{4}{2}$.