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?
Mickey and Donald choose a plate...
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- thephoenix
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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)
when bth choses diff plates there are 2C1 ways=2
tot=2+1=3(ways)
- sars72
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so, we consider the 3 white plates to be one entity?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)
- thephoenix
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yes as per question these are undistinguishable ...hence one entitysars72 wrote:so, we consider the 3 white plates to be one entity?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)
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.
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.