# Permutations question...

• Jan 26th 2009, 11:07 AM
Saibot
Permutations question...
In how many ways can 12 basketball players be assigned to six double rooms?

12!
__
(2!)^6

But i dont understand why its done that way.
• Jan 26th 2009, 11:27 AM
Plato
The number of rearragements of the string "AABBCC" is $\frac{6!}{(2!)^3}$.
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.
• Jan 26th 2009, 02:53 PM
Soroban
Hello, Saibot!

Quote:

In how many ways can 12 basketball players be assigned to six double rooms?

The answer is: . $\frac{12!}{(2!)^6}$

But i don't understand why it's done that way.

We could simply line up the twelve players: . $ABCD{E}FGHIJKL$
. .
There are: 12! permuations.

Then partition them in pairs: . $AB|CD|EF|GH|IJ|KL$

So that $A,B$ go to room #1 . . . $C,D$ go to room #2 . . . etc.

But among the $12!$ possible permutations of the twelve players,
. . there is a lot of duplication.

With our game plan, another permuation would be: . $BACD{E}FGHIJKL$

But this partitioning, $BA|CD|EF|GH|IJ|KL$
. . is identical to the first room assignment.
. . ( $AB$ is in #1, $CD$ 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: . $2^6$ possible reversals (each creating a duplication).

To eliminate the duplications, we divide by $2^6$

Therefore, the number of room assignments is: . $\frac{12!}{2^6}$

I hope all that is clear enough . . .

• Jan 26th 2009, 03:13 PM
Saibot
Thankks :D

very very gj :D

now i jsut have one moer question :'(