# Combinatorics counting problem

There are $$3n$$ male students and $$3n$$ female students. How many ways can they be divided into two groups of three, in such way that in each group has at least one male student and one female student.

I'd like to know if the following is correct: First I organize them in lines of $$2n$$, so: $$2n!\cdot 2n!$$. then, I create a third line with the rest, so in total: $$(2n!)^3$$. But now, I have created order inside the triplets, so to remove it I divide by $$(3!)^{2n}$$.

The final solution should be: $$\frac{(2n!)^3}{(3!)^{2n}}.$$

Is the solution above correct? Is the way of thinking correct?

• No this is not correct, since when $n=1$, it's not even an integer? I believe when $n=1$, the answer should be 6C3-1=19. I do not know the correct answer though – Gareth Ma Mar 28 at 12:28
• Also I do not understand your logic at all? – Gareth Ma Mar 28 at 12:29
• I don't understand your method. My suggestion: first show that there must be exactly $n$ groups of the form $MMF$ and exactly $n$ of the form $MFF$. Then populate each type separately and combine. – lulu Mar 28 at 12:37
• Should "two groups of three" instead be $2n$ groups of three? – RobPratt Mar 29 at 2:11

I think that the answer should be $$\frac{((3n)!)^2\binom{2n}{n}}{(2n)!2^{2n}}=\left(\frac{(3n)!}{2^{n}n!}\right)^{2}.$$ Explanation: we have two lines of $$3n$$ persons, one with males and another for females. Then we form $$2n$$ groups of $$3$$ persons each by taking $$1$$ male and $$1$$ female from the respective lines and then we fill the rest with the tails of the lines into $$\binom{2n}{n}$$ ways. Finally we divide by the number of permutations of the $$2n$$ groups, i.e. $$(2n)!$$ and by $$2$$ for each group (since we have a pair).

• we choose $n$ males who will be in their own groups, then divide the remaining $2n$ into pairs in $x$ ways. the same for the females, in $x$ ways. Then we pair the $n$ lone male with $n$ female pairs in $y$ ways. The same for the lone female and male pairs, $y$ ways. The answer is $(xy)^{2}$, always a square, but Yours is not? – Rezha Adrian Tanuharja Mar 28 at 13:15
• now our answer is equal:) – Rezha Adrian Tanuharja Mar 28 at 13:21
• For n=1, there are 3 male students: 1,2,3 and 3 female students: 4,5,6. the following are valid: 124,125,126,134,135,136,234,235,236,451,452,453,461,462,463,...,>9 For n=1 in your formula I get: 9. EDIT: A more detailed answer. sorry, had a mistake calculating. and thank you for you help!! – Tomer Attali Mar 28 at 21:52
• No, the number of possible divisions for $n=1$ is 9: 1) (124)(356), 2) (134)(256), 3) (234)(156), 4) (125)(346), 5) (135)(246), 6) (235)(146), 7) (126)(345), 8) (136)(245), 9) (236)(145). – Robert Z Mar 28 at 21:57
• I get confused by this alot. thank you for your help. – Tomer Attali Mar 28 at 21:59

Here is an alternative:

There will be $$n$$ groups containing $$1$$ male and $$2$$ females and the other $$n$$ group containing $$2$$ males and $$1$$ female.

The number of ways to divide the males into these $$2n$$ unordered groups is $$\frac{(3n)!}{2^{n}n!n!}$$.

The number of ways to divide the females into these $$2n$$ ordered group (yes ordered) is $$\frac{(3n)!}{2^{n}}$$.

Thus the number of ways to divide these males and females into group of $$3$$ with this requirement is $$\left(\frac{(3n)!}{2^{n}n!}\right)^{2}$$

• Nice answer! (+1) – Robert Z Mar 28 at 13:27
• What's the thinking behind having females be ordered? How does that determine the 1F/2M pairing I simply had them unordered, then said there were $(n!)^2$ ways to pair up the 1 M (resp 1F), 2 F (resp 2M) to form a group. – Calvin Lin Mar 28 at 13:51
• @CalvinLin exactly, I was just lazy to cancel the permutation by dividing by $(n!)^{2}$ only to multiply by it again when pairing male and female – Rezha Adrian Tanuharja Mar 28 at 15:11
• For n=1, there are 3 male students: 1,2,3 and 3 female students: 4,5,6. the following are valid: 124,125,126,134,135,136,234,235,236,451,452,453,461,462,463,...,>9 For n=1 your formula give me: 9. – Tomer Attali Mar 28 at 21:48
• @TomerAttali 136 and 452 should be counted as 1, 126 and 453 too and so on. we want to count the number of ways to divide into group of 3, not to count the number of distinct group of 3. Your way of counting should yield 18, divide it by 2 to get 9 – Rezha Adrian Tanuharja Mar 28 at 23:47