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Mickey and Donald choose a plate...

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Mickey and Donald choose a plate...

by pagalmes » Tue Jan 19, 2010 1:41 pm
From a pile of 4 plates, 3 undistinguishable whites plates and 1 blue plate, Mickey and Donald choose a plate. How many arrangements are there?

My answer would be : {Mickey | Donald} : {W, W}, {W, B}, {B, W}.

However, I can't manage to use the formulas to calculate it.
Is there a way to formalize that problem using formulas (permutation, combination)?

I would say that there is no importance of being Mickey or Donald choosing first.
But Donald having a white plate, and Mickey having a Blue plate {D:W, M:B} is different from {D:B, M:W}.

Any clue?
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by thephoenix » Thu Jan 28, 2010 8:09 pm
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)
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by sars72 » Thu Jan 28, 2010 8:20 pm
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?
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by thephoenix » Thu Jan 28, 2010 9:23 pm
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
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by regor60 » Fri Jan 29, 2010 8:42 am
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
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by pagalmes » Sat Jan 30, 2010 2:54 am
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.
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