1. ## probability..Car Seats!

A mini-van has two seats in front, a middle seat with spaces for three people, and a back seat with spaces for four people. Nine licensed drivers are going to ride in the van. One insists on sitting in the front seat, another insists on sitting in the middle seat, and a third insists on the sitting in the back seat. How many different seating arrangements satisfy everyone?

Please kindly give the proper explanation on how to resolve it...

2. Originally Posted by vikramtiwari
A mini-van has two seats in front, a middle seat with spaces for three people, and a back seat with spaces for four people. Nine licensed drivers are going to ride in the van. One insists on sitting in the front seat, another insists on sitting in the middle seat, and a third insists on the sitting in the back seat. How many different seating arrangements satisfy everyone?

Please kindly give the proper explanation on how to resolve it...

Use the pigeon hole principle. You can reverse engineer from my answer:

(2 x 6)(3 x 5 x 4)(4 x 3 x 2 x 1) = ......

(Then again, my batting average on combinatorial problems is about 0.3 so be warned )

3. Hello, vikramtiwari!

A mini-van has 2 seats in front, 3 seats in the middle, 4 seats in the back.
Nine licensed drivers are going to ride in the van. One insists on sitting
in a front seat, another on a middle seat, and a third on a back seat.
How many different seating arrangements satisfy everyone?

The one insisting on a front seat has 2 choices.
The one insisting on a middle seat has 3 choices.
The one insisting on a back seat has 4 choices.
. . Then the remaining 6 people can be seated in 6! ways.

Therefore, there are: . $(2\cdot3\cdot4)\cdot6! \;=\;\boxed{17,280}$ satisfactory seating arrangements.

4. Originally Posted by Soroban
Hello, vikramtiwari!

The one insisting on a front seat has 2 choices.
The one insisting on a middle seat has 3 choices.
The one insisting on a back seat has 4 choices.
. . Then the remaining 6 people can be seated in 6! ways.

Therefore, there are: . $(2\cdot3\cdot4)\cdot6! \;=\;\boxed{17,280}$ satisfactory seating arrangements.
So my batting average has risen (either that or yours has fallen!)