In how many ways can 12 basketball players be assigned to six double rooms?
The answer is
12!
__
(2!)^6
But i dont understand why its done that way.
The number of rearragements of the string "AABBCC" is $\displaystyle \frac{6!}{(2!)^3}$.
Because there are six letters, but three letters are repeat twice.
So we divide to account for that.
So think about a string "AABBCCDDEEFF". That is six rooms A-F, two to a room.
Then the team roster can correspond to any one of the arrangements of the 12-length string.
Compare your given answer with this.
Hello, Saibot!
In how many ways can 12 basketball players be assigned to six double rooms?
The answer is: .$\displaystyle \frac{12!}{(2!)^6}$
But i don't understand why it's done that way.
We could simply line up the twelve players: .$\displaystyle ABCD{E}FGHIJKL$
. . There are: 12! permuations.
Then partition them in pairs: .$\displaystyle AB|CD|EF|GH|IJ|KL$
So that $\displaystyle A,B$ go to room #1 . . . $\displaystyle C,D$ go to room #2 . . . etc.
But among the $\displaystyle 12!$ possible permutations of the twelve players,
. . there is a lot of duplication.
With our game plan, another permuation would be: .$\displaystyle BACD{E}FGHIJKL$
But this partitioning, $\displaystyle BA|CD|EF|GH|IJ|KL$
. . is identical to the first room assignment.
. . ($\displaystyle AB$ is in #1, $\displaystyle CD$ is in #2, etc.)
So if a pair is reversed, it results in a duplicated room assignment.
How many reversals are there?
Each of the six pairs can be reversed or not-reversed.
So there are: .$\displaystyle 2^6$ possible reversals (each creating a duplication).
To eliminate the duplications, we divide by $\displaystyle 2^6$
Therefore, the number of room assignments is: .$\displaystyle \frac{12!}{2^6}$
I hope all that is clear enough . . .