# Combinations problem,?

There are 20 students in a classroom, including person A and person B. The teacher splits up all 20 kids into 5 groups of 4 kids each.

I thought of this problem, and what are the total number of different groups there could be? Also, how many of these ways would have A and B on the same team? So then what's the probability that person A and person B will be on the same team?

My try for the first $\binom{20}{4} \binom{16}{4} \binom{12}{4} \binom{8}{4} \binom{4}{4}$

The second is not clear, any help?

• Your calculation seems off, assuming the teams are indistinguishable. You have to divide by the ways to permute the teams that are formed. – lulu Oct 19 '17 at 20:13
• To be clear: suppose you were making teams of $4$ from $8$ students. The answer is $\frac 12\times \binom 84$ as choosing $ABCD$ is the same as choosing $EFGH$. – lulu Oct 19 '17 at 20:14
• @lulu why not $4!$ ? – user373141 Oct 19 '17 at 20:17
• It's $5!$ as there are $5$ teams. – lulu Oct 19 '17 at 20:21

$\underline{Second\; problem}$

Place A and B together in any group, and note that this group has now become labelled.

Place $2$ more in this group in $\binom{18}2$ ways,
now we need to distribute $16$ in $4$ unlabelled groups,
so proceeding in exactly the same way as for the first, $\binom{16}4\binom{12}4\binom{8}4\binom44 / 4!$

You can now find out the probability.

Actually, if for problem two, you only need to find the probability, there is a much simpler way.

Place A anywhere. In A's group, 3 slots are vacant against 19 total vacant slots,
hence P(A and B are in the same group) = $\dfrac3{19}$

as you mentioned in the question the groups are the same. so your final answer is not correct.you need to divide the final answer by 5!.because the groups are the same but you are counting for example groups like this: ABCDE,ABCED,ACEDB etc so you need to divide the final answer

• what's about the second? – user373141 Oct 19 '17 at 20:51