# Thread: Extremely Simple Circle Probability Problem! :)

1. ## Extremely Simple Circle Problem! :)

How many ways can 8 boys be arranged in a circle if 3 boys cannot sit next to each other?

Well, I don't actually have the correct answer, but this is how I solved it:

Total Ways the Boys Can Be Arranged - 3 Boys Sitting Together = Ways The 3 Boys Cannot Sit Together

So, I did:

(8-1)! - (5! x 3!) = 4320 ways

I think this is wrong though. If someone would be able to solve it and explain the reasoning behind it, that would be awesome!

2. Hello, saintv!

How many ways can 8 boys be arranged in a circle
if 3 boys cannot sit next to each other?

This is how I solved it:

(Total ways boys can be arranged) - (3 boys together) .= .(Ways 3 boys are not together)

So, I did: .$\displaystyle (8-1)! - (5!\cdot 3!) \:=\: 4320$ ways

I think this is wrong though.
It depends on the interpretation of "the 3 boys are not together".

Suppose the eight boys are: {A, B, C, D, E, X, Y, Z}
. . and X, Y, Z must not sit in adjacent chairs.

If "the 3 boys are together" means $\displaystyle XYZ$ (in some order),
. . your solution is correct.
Note that this allows, for example: . X A Y Z B C D E
. . where two of them can be together.

If no two of those boys can be adjacent, a different approach is used.

Place the other 5 in a circle; there are: .$\displaystyle (5-1)!=24$ ways.
. . There are 5 spaces between these 5 boys.
Select 3 of these spaces to place X, Y, Z.
. . There are: .$\displaystyle _5P_3 = 60$ ways.

Therefore, there are: .$\displaystyle 24 \times 60 \,=\,1440$ ways.