Mickey and Donald choose a plate...

when bth choses white plates there is only one way or 1C1=1
when bth choses diff plates there are 2C1 ways=2
tot=2+1=3(ways)

thephoenix wrote:when bth choses white plates there is only one way or 1C1=1
when bth choses diff plates there are 2C1 ways=2
tot=2+1=3(ways)

so, we consider the 3 white plates to be one entity?

sars72 wrote:

thephoenix wrote:when bth choses white plates there is only one way or 1C1=1
when bth choses diff plates there are 2C1 ways=2
tot=2+1=3(ways)

so, we consider the 3 white plates to be one entity?

yes as per question these are undistinguishable ...hence one entity

Just don't assume that each of the 3 is equally likely...chances are 50% of getting WW, and 25% each of WB and BW

Thanks for the answers.
Here is a way of representing the problem:

There are 2 empty slots and enough white plates to fill them.

_ _

Thus, there are 2 white plate out of wich you have to choose blue plates:

W W

You can choose zero or 1 blue plate:

- from 2 white plates, choose zero blue plates (2C0 = 1)
- from 2 white plates, choose one blue plates (2C1 = 2)

Thus, the number of arrangements is 1 + 2 = 3.

Here is a more complex problem using the same logic:

you now have 4 persons, M,D, P, E that choose from a pile of 4 white plates and 2 blue plates. Again we can start with 4 white plates, and choosing the blue plates out of them:

- from 4 white plates, choose zero blue plates (4C0 = 1)
- from 4 white plates, choose one blue plates (4C1 = 4)
- from 4 white plates, choose zero blue plates (4C2 = 6)

4C0:
W W W W

4C1:
W W W B
W W B W
W B W W
B W W W

4C2:
W W B B
W B W B
B W W B
W B B W
B W B W
B B W W

Thus, you have 1 + 4 + 6 = 11 possible arrangements.