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suppose I have a single set of 9 elements {A, B, C, D, E, F, G, H, I} and that I need to create 3 groups of 3 elements each. I know that I can do:

9C3 * 6C3 * 3C3 and that is 1680 different combinations but in those combinations may appear {{A, B, C}, {D, E, F}, {G, H, I}} and {{D, E, F}, {G, H, I}, {A, B, C}}.

Which is the formula to calculate this? I have the notion that I need to divide by something by don't know by what.

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    $\begingroup$ let your triplets be $T_1, T_2, T_3$ in how many ways can you arrange $T_i$'s in a row? $\endgroup$
    – mm-aops
    May 28 '14 at 21:56
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Your reasoning shows that there are $\binom{9}{3}\binom{6}{3}\binom{3}{3}$ ways to divide our people into three teams of three each, one of the teams to wear blue uniforms,one to wear white, and one to wear red.

Every "nudist camp" division into teams gives rise to $3!$ divisions into uniformed teams. So if $N$ is the number of nudist camp divisions, then: $N=\frac 1{3!}\binom{9}{3}\binom{6}{3}\binom{3}{3}$.

Remark: Alternately, line up our people in order of age, or student number, or whatever. The first person can choose the rest of her team in $\binom{8}{2}$ ways. The first unchosen person can then choose the rest of her team in $\binom{5}{2}$ ways. So the number of ways is $\binom{8}{2}\binom{5}{2}$.

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  • $\begingroup$ Yeah, that would be if the order of blue, white and red teams matter but in my question it does not, {{A, B, C}, {D, E, F}, {G, H, I}} and {{D, E, F}, {G, H, I}, {A, B, C}} would be considered same answers. Only the inner groups have to be different. $\endgroup$ May 28 '14 at 22:10
  • $\begingroup$ The uniformed count was just a prelude to counting $N$, the actual number you are looking for. The post shows that $(N)(3!)$ is the number you got, so to get $N$ we divide your expression by $3!$. $\endgroup$ May 28 '14 at 22:13
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There are $9 \choose 3$ ways to select the $3$ letters for the first group, then there are $6 \choose 3$ ways to choose $3$ letters to form the second group. And the remaining $3$ letters will automatically be in the group with the first six chosen letters. So the answer is: $9 \choose 3$$\times$ $6 \choose 3$ ways.

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  • $\begingroup$ But when the order of the groups does not matter you have over counted. Each arrangement is one of a group of $3!$ equivalent arrangements. So the answer is then: $\frac 1{3!} \,^9\mathrm{C}_3\,^6\mathrm{C}_3$ $\endgroup$ May 28 '14 at 23:18

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