In how many ways we can split set$ \{a_{1},..,a_{9}, b_{1},.., b_{9}, c_{1} ,.., c_{9}\}$ into 9 set of shape $\{ a_{i}, b_{j}, c_{k} \} $

Question

In how many ways we can split set$ \{a_{1},..,a_{9}, b_{1},.., b_{9}, c_{1} ,.., c_{9}\}$ into 9 set of shape $\{ a_{i}, b_{j}, c_{k} \}$?

Answer 1

For choosing the first "team" we have $9*9*9$ ways to pick it. The second team has $8*8*8$ ways. Following the pattern, we get to a number of arrangements totaling $(9!)^3$

However, we are over counting the total number of ways by $9!$ as we don't care about the order the teams were picked, it only matters which elements are on which team.

Therefore, our answer is $\dfrac{(9!)^3}{9!} = (9!)^2$

Generalizing this, if we have the set $\{a_{1,1}, a_{1,2}, \dots a_{1,k}, a_{2,1}, a_{2,2},\dots a_{2,k}, \dots a_{n,1}, a_{n,2}, \dots a_{n,k}\}$

(so if instead of $3$ letters we have $n$ letters, and instead of having $9$ of each letter we have $k$ of each letter)

and we want to split this into $k$ sets of shape $\{a_{1,j1},a_{2,j2}\dots a_{n,jn}\} $ the number of ways to arrange them into teams is $$\frac{(k!)^n}{(k!)} = (k!)^{n-1} $$

Answer 2

We need to group them into tuples. Ok so $9^3$ ways to get the first tuple, then $8^3$ ways to get the second tuple and so on for a total of $9^3*8^3*7^3*...*1$?. But then we don't care about order.so divide by $9!$!! And that gives :$\frac{9^3*8^3*7^3*...*1^3}{9!}=131681894400$

Answer 3

Another way to get the answer $(9!)^2$ is to imagine the $a_i$'s already written, in order ($a_1,a_2,\dots,a_9$) in a row. A splitting of the sort you asked about amounts to a way to write a row of $b$'s underneath this row of $a$'s and then write a row of $c$'s underneath that. (The sets in your splitting will be the columns of this $3\times9$ matrix.) There are $9!$ ways to order the $b$'s and $9!$ ways to order the $c$'s, and therefore $(9!)^2 options altogether.