Hello, justinc89!

Your factorial is incorrect.

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

You should use 3 instead of 3!.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. ?

To divide them into (2,3,3), there are: . 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: . ways for (2,3,3).

To divide them into (2,2,4), there are: . ways.

The decision is: Who gets the "4"? . . . There are 3 choices.

So there are: . ways for (2,2,4).

Therefore, there are: . ways.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Their answer iswaytoo big.

Withnorestrictions, there are only: . ways.