# Homework Help: Probability: Arrangements in Circle

1. Sep 10, 2004

### futb0l

Ten children are arranging themselves in a circle. Calculate the number of ways this can be done if three particular children are NOT to be next to each other.

I did it and got... 9! - 7!3!
However, I got it wrong on the test - I am pretty sure my method is correct.

2. Sep 10, 2004

### HallsofIvy

That bit about "three particular children are NOT to be next to each other complicates things!

Try this. Take one of the 3 "particular children" and place him/her wherever you like.
There are 9 children left to place but only 7 of those can be next to the one already place. There are 7 choices for the child to place to the left and 7 for the child to be place to the right: 7*7 possibilities so far. We have place three children and have 7 left. Now continue to the left (clockwise). Since the child just place is NOT one of the "terrible three", we can place any of the remaining 7 children there. That now gives 7*7*7= 73 choices.
We now have six children left to place but now it gets complicated. IF the child just placed was one of the trouble makers (I'm not sure WHY they can't be placed next to each other but they are certainly making trouble for us!) then only 5 of the remaining 6 can be placed next to him/her: 5 choices. On the other hand, if it was ot, then any of the 6 can be placed there. This breaks into two parts now:
73(5*... + 6*...) and we have to follow both "paths".
Assuming the fourth child placed was one of the three: we place one of the 5 that can be beside him/her and now we can place the remaining 5 children (including that one trouble maker) as we please (remember that we placed a "non-trouble maker" to the right of the first child so we won't "run into" that child) That gives a total number of choices of 73*5*5*4*3*2*1.
Assuming the fourth child was NOT one of the three, we can place any of the remaining 6 next to him/her (my head is starting to hurt!) . For this path, so far, we have 73*6*6 but now we have to "branch" depending upon whether this lastest child was one of the three.
I'm going to stop now. I think I have given you enough to work out the various "paths" neccessary. After you have worked out the number of choices on each "path", add them all together.