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?
7 answers
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There are 3 options (boxes) for each of the 4 balls, so the number of ways is 3^4 = 81
!@#$%^& it this was wrong
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.
(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.
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 $1 + 4 + 3 + 6 = \boxed{14}$.
(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 $1 + 4 + 3 + 6 = \boxed{14}$.
thats copied straight from alcumus
You guys are cheating. Ask in the alcumus forum or the message board
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