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.

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- Jan 26th 2009, 10:07 AMSaibotPermutations question...
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. - Jan 26th 2009, 10:27 AMPlato
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. - Jan 26th 2009, 01:53 PMSoroban
Hello, Saibot!

Quote:

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 . . .

- Jan 26th 2009, 02:13 PMSaibot
Thankks :D

very very gj :D

now i jsut have one moer question :'(