# Thread: permutations - too many balls!

1. ## permutations - too many balls!

Hi folks,

I thougth I could do these permutation questions until now!

I have 6 balls, 3 red, 1 blue, 1 brown, and 1 green. In how many ways can I arrange them in a line? In how many of these arrangements will no two red balls come together?

The first part is fairly standard i.e arrangements of n things with p things of the same kind gives n! / p! arrangements:

6! / 3! = 6 x 5 x 4 = 120.

The second part would be easy too if there were only 2 red balls. We would consider the two like objects as a single object so there would be 5 objects arranged in 5! ways and since the two red balls can be arranged in 2! ways amoung themselves, we would have 2.5! arrangements with the two balls together, so

120 (total perms) - 2.5! (arrangements in which the two red balls are together) = arrangements in which the two red balls are not together.

But this 3rd red ball is the problem! I tried the obvious, namely joining the 3 red balls to produce 4 entities i.e. 4! arrangements with 3! arrangements of the 3 red balls so that:

total perms - 4!3! = arrangements with no three red balls together

but then I noticed that I am asked to find how many arrangements with no two red balls together. I also not clear on whether total perms is 6! or 6!/3!
6!/3! - 4!/3! is a negative so something is definitely wrong!

Any ideas?

2. Originally Posted by s_ingram
I have 6 balls, 3 red, 1 blue, 1 brown, and 1 green. In how many of these arrangements will no two red balls come together?
The three odd balls make four possible places for the red balls:_Bl_Br_G_.
So there $\binom{4}{3}=4$ ways to place the red balls.
But the string _Bl_Br_G_ can be arranged in $3!=6$ way.
So how many is that total?

3. Yep, that's the right answer. Seems I was on the wrong track completely. Thanks again.