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Permutations/combinations question:

Number of ways 24 students of 12 boys and 12 girls, how many ways can be separated into groups of 3 if all groups have to be mixed gender?

I thought I had the answer when I did this:

  • Take 8 boys, and put them each into one group. This can be done in 8! ways.
  • Take 8 girls, and put them each into one group. Once again, 8! ways.
  • There are 4 girls and 4 girls left, and you can put either a boy or a girl in the last position, so that’s 8!x8!x8!

But, This can’t be right because it doesn’t the times when, for example, two of the boys from the group of 8 are in the same group. Can anyone help me out?

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  • $\begingroup$ Notice that you want to create four groups with two boys and one girl and four groups with two girls and one boy. $\endgroup$ – N. F. Taussig Sep 2 '18 at 10:57
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You're idea is right, but it needs some refinement. First you have to select the $8$ boys in the first step and similarly the $8$ girls in the second one. Moreover for the last step we have $8!$ ways to do it. Indeed the first boy can be placed in $8$ groups, the next one in $7$ and so on.

Finally to avoid double counting you have to divide by $(2!)^8$. This is because each group has $2$ boys (or girls) and each of it can be used in the first step of creating the very same group.

Additionally you want to divide by $8!$, as I assume the group are undistinguishable. I believe that having placing boy $1,2$ and girl $1$ in the first group is same as placing them in the second one. Thus the final answer is:

$$\frac{\binom{12}{8} \cdot 8! \cdot \binom{12}{8} \cdot 8! \cdot 8!}{(2!)^8 \cdot 8!}$$

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  • $\begingroup$ I don’t quite understand why you divide by (2!)^8? $\endgroup$ – M. Weate Sep 2 '18 at 11:15
  • $\begingroup$ @M.Weate As I said because we have two boys in a group, there are two ways to create the group. Each of them can be chosen in the first step. Thus we are double counting each cofiguration $\endgroup$ – Stefan4024 Sep 2 '18 at 11:50
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    $\begingroup$ @N.F.Taussig I was trying to add up the solution of the OP. In his solutiong he is giving numbers to the groups. Indeed it can be seen from the construction. We choose $\binom{12}{8}$ boys and we have $8!$ ways to distribute them in the eight groups. If the groups weren't numbered then the $8!$ factor is redundant. Thus after completing the process we have triplets of people in numbered groups. However the number of the groups doesn't make any difference, so we divide by $8!$. Furthermore both our answers agree. $\endgroup$ – Stefan4024 Sep 3 '18 at 11:41
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Think of it this way. Consider a 3x8 grid. Every such grouping can be obtained by permuting the 12 girls among the top row and first 4 places of the second row and permuting the boys among the bottom row and last 4 places of the second row. Then the groups are the columns of this grid.

So there are clearly $12! \times 12!$ ways to do this. Then we ask ourselves: how many permutations lead to the same grouping? Or equivalently, how many ways are there to permute a grouping without changing it? Then we simply divide $12! \times 12!$ by this number to get the number of groupings.

Spoiler:

$$\frac{12!\times 12!}{4! \times 4! \times (2!)^8}.$$

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  • $\begingroup$ I don’t quite understand why you divide by (2!)^8? $\endgroup$ – M. Weate Sep 2 '18 at 11:17
  • $\begingroup$ @M.Weate How many ways are there to permute the boxes around without changing the grouping? If I swap the two girls in the first group, that doesn't change the groupings. If I swap the first two columns, that doesn't change the groupings. Etc. $\endgroup$ – Trevor Gunn Sep 2 '18 at 11:19
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Observe that since each of the eight groups of three people is mixed gender, there must be four groups with two boys and one girl and four groups with two girls and one boy.

Selecting four groups with two boys and one girl: The boys for the "first" group can be selected in $\binom{12}{2}$ ways and the girl in that group can be selected in $\binom{12}{1}$ ways. The boys for the "second" group can be selected in $\binom{10}{2}$ ways and the girl in that group can be selected in $\binom{11}{1}$ ways. The boys for the "third" group can be selected in $\binom{8}{2}$ ways and the girl in that group can be selected in $\binom{10}{1}$ way. The boys in the "fourth" group can be selected in $\binom{6}{2}$ ways and the girl in that group can be selected in $\binom{9}{1}$ ways. However, the order in which we select these four groups does not matter, so we must divide by the $4!$ orders in which we could select the same four groups. Hence, the number of such selections is $$\frac{1}{4!}\binom{12}{2}\binom{12}{1}\binom{10}{2}\binom{11}{1}\binom{8}{2}\binom{10}{1}\binom{6}{2}\binom{9}{1}$$

Selecting four groups with two girls and one boy: There are eight girls and four boys remaining. There are $\binom{4}{1}$ ways to select the boy for the "first" group and $\binom{8}{2}$ ways to select the girls in that group. There are $\binom{3}{1}$ ways to select the boy in the "second" group and $\binom{6}{2}$ ways to select the girls in that group. There are $\binom{2}{1}$ ways to select the boy in the "third" group and $\binom{4}{2}$ ways to select the girls in that group. The remaining boy and girls must comprise the "fourth" group. However, the order in which the groups are selected does not matter, so we must divide by the $4!$ orders in which we could select the same four groups. Hence, the number of such selections is $$\frac{1}{4!}\binom{4}{1}\binom{8}{2}\binom{3}{1}\binom{6}{2}\binom{2}{1}\binom{4}{2}\binom{1}{1}\binom{2}{2}$$

Total: Multiplying the two results and simplifying yields $$\frac{12!12!}{4!4!2!^8}$$

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There will be $4$ groups with one boy and $4$ groups with two boys. Four single boys can be selected in ${12\choose 4}=495$ ways; then by letting the oldest remaining boy choose his mate in turn the two-boy groups can be formed in $7\cdot5\cdot 3=105$ ways. Same thing for the girls. Finally we can combine the singles with the pairs of the other sex in $(4!)^2$ ways. It follows that there are $$\bigl(495\cdot105\cdot 4!\bigr)^2=1\,556\,006\,760\,000$$ admissible configurations (same as in N.F. Taussig's answer).

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