Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

10 boys and 2 girls are divided into 3 groups of 4 each. The probability that the girls will be in different groups is?

share|cite|improve this question
...low?${}{}{}$ – Shahar Aug 28 '14 at 1:08
up vote 5 down vote accepted

A simple way is to suppose that person A has been assigned to Group 1. Note that $3$ of the remaining $11$ will be assigned to Group 1. The probability that person B is one of them is $\frac{3}{11}$. So the probability that A and B end up in different groups is $\frac{8}{11}$.

There are more elaborate combinatorial arguments.

Remark: At the request of OP, here is a more complicated argument. Without changing the probabilities, we may assume that the groups are named groups, say Groups 1, 2, and 3.

The number of (equally likely) ways to assign people to named groups is $\frac{12!}{4!4!4!}$. One way of seeing this is that there are $\binom{12}{4}$ ways to decide who goes into Group 1, and for each of these there are $\binom{8}{4}$ ways to decide who goes into Group 2.

Now we could directly count the number of ways to assign A and B to different groups, or do it indirectly by counting the number of ways to assign A and B to the same group. We do the second.

There are $\binom{3}{1}$ ways to choose the common group. For each of these, there are $\binom{10}{2}$ ways to choose the groupmates of A and B. And there are $\binom{8}{4}$ ways to decide on the members of the first unused group, for a total of $\binom{3}{1}\binom{10}{2}\binom{8}{4}$.

Divide by $\frac{12!}{(4!)^3}$ to find the probability A and B end up in the same group. Fairly quickly, the expression simplifies to $\frac{3}{11}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.