Asked by Epic
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?
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?
Answers
Answered by
Hi
bump
Answered by
Solstice
There are 3 options (boxes) for each of the 4 balls, so the number of ways is 3^4 = 81
Answered by
jeff
!@#$%^& it this was wrong
Answered by
Naomi
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.
Answered by
Staragon
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 $1 + 4 + 3 + 6 = \boxed{14}$.
Answered by
poo
thats copied straight from alcumus
Answered by
Cheater
You guys are cheating. Ask in the alcumus forum or the message board
There are no AI answers yet. The ability to request AI answers is coming soon!