# Permutations and Combinations exam question

Before I proceed with my queries I think it's best to present the question at hand.

A class consisting of 4 males and 12 females in randomly divided into 4 groups of 4. What is the probability that each group consists of a male and three females?

Now the given answer for this was $\frac{64}{455}$.

The closest I could possibly get to that was $455$.

Whenever I come across these types of questions I generally categorise it like so:

• Does order matter?
• Yes: then it is a permutation (% increases)
• No: then it is a combination (% decreases)
• Does it allow repetitions? - Make use of respective formulae

So reading the question I will immediately think, "No, order does not matter" which I find a bit hard to explain but Male Student A may be in the first group and Male Student B in the second group OR Vice-Versa (hence why I would consider it a combination..).

So once that is out of the way, I proceed and think, "Yes, there may be repetitions". For consistencys sake, Male Student A may be in the first group twice.

Now that I got that chunk out of the way I pray to the seven Gods that I got the previous two right (even though it's purely logic, certain question try to trick you), and hence begin working.

Pardon if the following notation is incorrect: $$C\binom{16}{4}$$

I tried the following: $$\therefore \frac{16!}{12!\times 4!}\div 4 = 455$$

And then: $$\therefore \frac{16!}{12!\times 4!}\div 4! = \frac{6}{455}$$

However the answer presented in my answer sheet is: $$\frac{64}{455}$$

Let's look at the combinations, and we'll do it one group at a time. We are looking for the probability that each of the four groups has exactly one boy and three girls.

For the first group, we are choosing $4$ people out of $16$. One is a boy, out of $4$ possibilities. Three are girls, out of $12$ possibilities. The choices of boy and girls are independent, so we can multiply their counts. Therefore, the probability that the first group has one boy and three girls is

$$\frac{{4 \choose 1}{12 \choose 3}}{{16 \choose 4}}=\frac{44}{91}$$

Now for the second group. After the first group is chosen suitably, we have to choose $1$ boy out of $3$ remaining and $3$ girls out of $9$ remaining. But the total number of possible groups is $4$ people out of $12$ remaining. Therefore, given that the first group was successful, the probability that the second group is successfully chosen is

$$\frac{{3 \choose 1}{9 \choose 3}}{{12 \choose 4}}=\frac{28}{55}$$

Now for the third group. After the first two groups are chosen suitably, we have to choose $1$ boy out of $2$ remaining and $3$ girls out of $6$ remaining. But the total number of possible groups is $4$ people out of $8$ remaining. Therefore, given that the first and second groups were successful, the probability that the third group is successfully chosen is

$$\frac{{2 \choose 1}{6 \choose 3}}{{8 \choose 4}}=\frac{4}{7}$$

If the first three groups are correct, so is the fourth. Therefore, the total probability of success with all four groups is

$$\frac{44}{91}\cdot\frac{28}{55}\cdot\frac{4}{7}=\frac{64}{455}$$

Or, if you want the final probability as one large calculation,

$$\frac{{4 \choose 1}{12 \choose 3}}{{16 \choose 4}} \cdot \frac{{3 \choose 1}{9 \choose 3}}{{12 \choose 4}} \cdot \frac{{2 \choose 1}{6 \choose 3}}{{8 \choose 4}} =\frac{64}{455}$$

• Such a great answer. Thanks for taking the time to explain all of this. Makes perfect sense. Apr 26 '15 at 18:08