1. ## seating placement

Hello, I thought this would of been easy but I am second guessing myself.

Frank, Tom, Jim and 6 other boys are randomly seated in a row. What
is the probability that neither Frank nor Tom sits next to Jim?

Here is what I came up with. However, I am not confident in it, because it seems rather low.

(6! x 9)/(9!) = 0.0178... or 1.78%

2. Calculate the probability that at least one of Frank or Tom does sit next to Jim.

If Frank sits next to Jim, then they can be arranged 2 ways, and considered as a unit. There are 2x8! possible arrangements in this case.

If Tom sits next to Jim, then the same is true. There are 2x8! possible arrangements.

If Frank and Tom both sit next to Jim, then the three of them can be arranged 2 ways (since Jim is always in the middle), and considered as a unit. There are 2x7! possible arrangements.

So the total is 2x8! + 2x8! - 2x7! = 4x8! - 2x7!

There are 9! arrangements in general.

So the probability of at least one of Frank or Tom sitting next to Jim is
$\displaystyle \frac{4(8!) - 2(7!)}{9!}$

And the probability that neither of them sits next to Jim is

$\displaystyle 1 - \frac{4(8!) - 2(7!)}{9!}$

3. Hello, chrisc!

Frank, Tom, Jim and 6 other boys are randomly seated in a row.
What is the probability that neither Frank nor Tom sits next to Jim?
I though I would solve this "head-on" . . .

There are: .$\displaystyle 9! \:=\:362,880$ possible seatings.

There are two cases to consider:
. . [1] Jim occupies an end seat.
. . [2]Jim does not occupy an end seat.

[1] Jim occupies an end seat.
First, there is a choice of 2 end seats he can occupy.

We have: . $\displaystyle J \;\_\;\_\;\;\_\;\_\;\_\;\;\_\;\_\;\_$

The second seat must not be taken by Frank or Tom.
. . It is taken by one of the other 6 boys.
Then the other seven boys can be seated in 7! ways.

Hence, there are: .$\displaystyle 2 \times 6 \times 7! \:=\:60,480$ ways with Jim in an end seat.

[2] Jim occupies an "inner" seat.
He has a choice of 7 inner seats.

We have: .$\displaystyle \_\;J\;\_\;\;\_\;\_\;\_\;\;\_\;\_\;\_$

The left seat must not be taken by Frank or Tom.
. . It is taken by one of the other 6 boys.
The right seat must not be taken by Frank or Tom.
. . It is taken by one of the other 5 boys.
Then the other six boys can be seated in 6! ways.

Hence, there are: .$\displaystyle 7 \times 6 \times 5 \times 6! \:=\:151,200$ ways with Jim in an inner seat.

Then there are: .$\displaystyle 60,480 + 151,200 \;=\;211,680 \text{ ways}$

Therefore, the probability is: .$\displaystyle \frac{211,680}{362,880} \;=\;\boxed{\frac{7}{12}}$

We agree, icemanfan!
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