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?
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
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.