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How many ways are there to put 4 balls in 3 boxes if two balls are indistinguishably green, two are indistinguishably red, and the boxes are distinguishable?

 Dec 16, 2017
 #1
avatar+99109 
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How many ways are there to put 4 balls in 3 boxes if two balls are indistinguishably green, two are indistinguishably red, and the boxes are distinguishable?

 

This is what I think.

A)  All in one box = 4 ways

B)  1 in a box and 3 in another box = 4*2 * 3 = 24 (choose a box, choose a colour, choose a second box)

C) 2 of one colour in a box and 2 of the other colour in another box

     (chose a box, chose a colour, chose another box)/2 = 4*2*3/2 = 24/2 =12

     OR     chose a box for the reds * chose a box for the greens = 4*3=12

D) RG in 2 of the boxes. 

      choose a box, chose another box divide by 2 =  4*3/2=12/2 =6

E) One in each box = 4!/(2!2!) = 6

 

4+24+12+6+6=52 ways

 Dec 16, 2017
 #2
avatar+474 
+3

Thank you, but that is incorrect.

 

Solution I found: 

We will consider this as a composite of two problems with two indistinguishable balls and 3 distinguishable boxes. For two indistinguishable green balls, we can place the balls in a box together or in separate boxes. There are 3 options to arrange them together (in box 1, 2, or 3) and 3 options for placing them separately (nothing in box 1, 2, or 3). So there are 6 ways to arrange the indistinguishable green balls. By the same reasoning, there are 6 ways to arrange the indistinguishable red balls, for a total of 6*6=36 arrangements of the 4 balls.

RektTheNoob  Dec 18, 2017
edited by RektTheNoob  Dec 18, 2017
 #3
avatar+99109 
0

I am sorry, I misread the question. I thought there were 4 boxes.

 Dec 18, 2017

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