How many ways are there to put 4 balls in 3 boxes if the balls are distinguishable but the boxes are not?
How many ways are there to put 4 balls in 3 boxes if two balls are indistinguishably white, two are indistinguishably black, and the boxes are distinguishable?
Without regard to the distinguishability of the balls, they can be organized into groups of the following:$$(4,0,0),(3,1,0),(2,2,0),(2,1,1).$$Now we consider the distinguishability of the balls in each of these options.
(4,0,0): There is only $1$ way to do this (since the boxes are indistinguishable).
(3,1,0): There are $4$ options: we must pick the ball which goes into a box by itself.
(2,2,0): There are 4 choose 2 = 6 ways to choose the balls for the first box, and the remaining goes in the second box. However, the two pairs of balls are interchangeable, so we must divide by 2 to get $6 / 2 = 3$ arrangements.
(2,1,1): There are 4 choose 2 = 6 options for picking the two balls to go in one box, and each of the other two balls goes into its own box.
The total number of arrangements is $1 + 4 + 3 + 6 =14.
thats copied straight from alcumus
You guys are cheating. Ask in the alcumus forum or the message board
Without regard to the distinguishability of the balls, they can be organized into groups of the following:$$(4,0,0),(3,1,0),(2,2,0),(2,1,1).$$Now we consider the distinguishability of the balls in each of these options.
(4,0,0): There is only $1$ way to do this (since the boxes are indistinguishable).
(3,1,0): There are $4$ options: we must pick the ball which goes into a box by itself.
(2,2,0): There are $\binom{4}{2} = 6$ ways to choose the balls for the first box, and the remaining go in the second box. However, the two pairs of balls are interchangeable, so we must divide by 2 to get $6 / 2 = 3$ arrangements.
(2,1,1): There are $\binom{4}{2} = 6$ options for picking the two balls to go in one box, and each of the other two balls goes into its own box.
The total number of arrangements is $1 + 4 + 3 + 6 = \boxed{14}$.