Permutations again

**Misa-Campo** · July 12th 2008, 12:23 PM

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?

**Plato** · July 12th 2008, 01:32 PM

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 $\binom {4} {2} = 6$ 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!).

**particlejohn** · July 12th 2008, 01:43 PM

Shouldn't it be $4 \binom{3}{2} \binom{4}{2} = 72$ ? Because you fix the $2$ people in the back seat and the $1$ person in the front seat?

**janvdl** · July 12th 2008, 02:25 PM

If you are getting help, don't scoff at other members or be rude. Thanks.

**Plato** · July 12th 2008, 02:26 PM

Originally Posted by particlejohn

Shouldn't it be $4 \binom{3}{2} \binom{4}{2} = 72$ ? Because you fix the $2$ people in the back seat and the $1$ 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. $\binom {4} {2}$ and two ways to arrange those two. At this point we have four people remaining for the back seat, $4!$ .

**particlejohn** · July 12th 2008, 02:31 PM

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. $\binom {4} {2}$ and two ways to arrange those two. At this point we have four people remaining for the back seat, $4!$ .

Or could you fix the $2$ particular people in the back seat? That means there are $5$ people left. $1$ must be the driver. So we have $4$ people from which we can choose the other $2$ for the front seat. Same answer: $2\cdot 4! \binom{4}{2}$ .

**Misa-Campo** · July 12th 2008, 02:36 PM

well sorry but i had no intention, was just being straightforward thats all, ill be more polite next time lol

**Plato** · July 12th 2008, 02:39 PM

Originally Posted by particlejohn

Or could you fix the $2$ particular people in the back seat? That means there are $5$ people left. $1$ must be the driver. So we have $4$ people from which we can choose the other $2$ for the front seat. Same answer: $2\cdot 4! \binom{4}{2}$ .

That works. You start with the back seat, I start with the front.

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