You have $3$ black and $2$ red balls. What is the probability that you can arrange them such that no black balls are placed consecutively?

I thought about it as being $$1- P(3~\text{black balls are next to each other}) – P(2~\text{black balls are next to each other})$$ But, $$P(3~\text{black balls are next to each other}) = \frac{(3 \cdot 6 \cdot 2)}{5\text{C}3} = \frac{18}{5}$$ which is clearly wrong as there are $3$ positions out of $5$ where $3$ balls can be placed together and these $3$ balls can be arranged in $6$ ways each time, with the other $2$ balls being arranged in $2!$ ways.

Cheers.

## Best Answer

There is no particular need to order the balls, as the question deals with a final outcome of black and red balls, with no differences in how the balls are permuted within their own color groups. So we need only have our sample space be the number of ways to place 2 red balls and 3 black balls into 5 slots, with no counting of the permutations within color groups. This is just $$\binom{5}{3}$$ since, in each case, there is only 1 way to fill the remaining 2 slots with 2 red balls (Don't worry about the permutations of the red balls. The final results cannot be discerned from each other). Now, as you correctly ascertained, there are only 3 ways to place 3 black balls consecutively in 5 slots. However, if we place 2 black balls consecutively, there are then 2 additional slots into which we can place the remaining black ball in each case where the 2 balls lie on the end (the slot adjacent to the two consecutive balls cannot be used as we have already counted the cases of three consecutive black balls), and 1 additional slot if the two consecutive balls lie in the middle. So the two end cases give us 4 unwanted positions, while the two middle cases give us only 2 unwanted positions. So there are 6 unwanted positions where only two black balls are consecutive.

Thus, there are $$\binom{5}{3} - 3 - 6$$ ways to place the balls so that no black balls are consecutive. The probability of this event is then $$\frac{\binom{5}{3} - 3 - 6}{\binom{5}{3}}$$ the size of the event (the number of favorable possibilities) over the size of the sample space (the number of all possibilities under consideration).

It is not incorrect to include the permutations within color groups of course, it is just another way of listing the individual events, and gives us a larger sample space. We will, however, get the same result: in this case, the method just adds unnecessary factors that cancel anyway. With permutations of $3!$ for the black balls and $2!$ for the red balls, the total sample space of events now has a size of $$\binom{5}{3}3!2!$$ and the number of favorable events (no consecutive black balls) is then $$\binom{5}{3}3!2! - 3\cdot 3!2! - 6\cdot 2!3!$$ Our probability will therefore be $$\frac{\binom{5}{3}3!2! - 3\cdot 3!2! - 6\cdot 2!3!}{\binom{5}{3}3!2!}$$ which you can see reduces to the same ratio.

This is actually a good technique for checking your own probability work, since there is rarely a way to check empirically. Just try two different ways to count the events in question and see if you get the same probability.