CPhill

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 #9
avatar+130555 
+10

We have two tasks here....choose the boxes....choose the b***s to go into those boxes...

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4 2 0

We have 3  ways to choose the first box......and we want to put 4 of the 6 b***s into it

3C(6,4)  = 45

Then, out of the remaning 2 boxes, we want to choose one of the two to put the remaining two b***s into

So 45 x 2 = 90

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4 1 1

Agian we have 3 boxes to choose which 4 b***s out of 6 we put into it

So 3C(6,4)= 45 ways

And we have two ways to select the next box and two ways to select which ball goes into that box...

The last box and ball are defaulted

So we have

2C(2.,1) = 4 ways

So 45 x 4 = 180

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3 3 0

3 ways to choose the first box......and we want to choose 3 of 6 b***s to put into it

3C(6,3) = 60 ways

Then, we have 2 ways to choose the next box to place the remaining three b***s into

So 2 x 60  = 120 ways

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3 2 1

3 ways to choose the first box....choose 3 of 6 b***s to put into it   = 60 ways  

2 ways to select the next box and we want to choose 2 of the 3 remaining b***s to put into it

2C(3.2) = 6

Again, the last box and ball are defaulted

So   60 x 6 = 360 ways

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2 2 2

3 ways to select the first box .... choose 2 of 6 b***s to go into it  = 3C(6,2) = 45

2 ways to pick the next box and choose 2 of the remaining 4 b***s to put into it

2C(4,2)  = 12

The last box and remaig two b***s are defaulted

So 45 x 12 = 540 ways

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Anyway, that's what I get......!!!

 

  

21.03.2015