1. ## number of ways

The question is:

In how many ways can 8 different CD's be distributed to 3 people so that each person receives at least 2 CD's?

My solution:

(8C2 * 6C3 * 3C3) * 3! = 3360

(number of ways that two people get 3 CD's and one person gets 2 CD's)

(8C2 * 6C2 * 4C4) * 3! = 2520

(number of ways that two people get 2 CD's and one people gets four CD's)

3360+2520= 5880

However, the correct answer is 6720. What have I done wrong? Any help is very much appreciated. Thanks

2. Hello, justinc89!

. . And I don't agree with their answer . . .

In how many ways can 8 different CD's be distributed to 3 people,
so that each person receives at least 2 CD's?

My solution:

(8C2 * 6C3 * 3C3) * 3! = 3360
. .
(number of ways that two people get 3 CD's and one person gets 2 CD's)

(8C2 * 6C2 * 4C4) * 3! = 2520
. .
(number of ways that two people get 2 CD's and one people gets four CD's)

3360 + 2520 = 5880

However, the correct answer is 6720. ?
You should use 3 instead of 3!.

To divide them into (2,3,3), there are: .$\displaystyle {8\choose2}{6\choose3}{3\choose3} \:=\:560$ ways.
But there are not 3! permutations; there are only 3.
. . [The decision is: Who gets the "2"? . . . There are 3 choices.)
So there are: .$\displaystyle 3 \times 560 \:=\:1680$ ways for (2,3,3).

To divide them into (2,2,4), there are: .$\displaystyle {8\choose2}{6\choose2}{4\choose4} \:=\:420$ ways.
The decision is: Who gets the "4"? . . . There are 3 choices.
So there are: .$\displaystyle 3 \times 420 \:=\:1260$ ways for (2,2,4).

Therefore, there are: .$\displaystyle 1680 + 1260 \:=\:\boxed{2940}$ ways.

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Their answer is way too big.

With no restrictions, there are only: .$\displaystyle 3^8 \:=\:6561$ ways.

3. Thanks a lot, I figured the answer key was wrong.