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Distinguishibility

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How many ways are there to put 6 b***s in 3 boxes if the b***s are distinguishable but the boxes are indistinguishable?

Jun 21, 2017

#1
+17747
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How many ways can you put 6 balls in 3 boxes if the balls are distinguishable but the boxes are not?

I believe that the answer is 207.

The balls can be arranged as:

-- 6 balls in one box and none in the other two                                                         6 - 0 - 0

-- 5 balls in one box and 1 in one of the other two boxes                                         5 - 1 - 0

-- 4 balls in one box and 2 in one of the other two boxes                                         4 - 2 - 0

-- 3 balls in one box and 3 in one of the other two boxes                                         3 - 3 - 0

-- 4 balls in one box and 1 ball in another box and 1 ball in the third box                 4 - 1- 1

-- 3 balls in one box and 2 balls in another box and 1 ball in the third box               3 - 2- 1

-- 2 balls in one box and 2 balls in another box and 2 balls in the third box.             2 - 2- 2

Now, for each of the possibilities.

6 - 0 - 0:  There is only 1 way to do this [ C(6,6) = 1 ]

5 - 1 - 0:  There are 6 ways to do this [ C(6,5) = 6 ]

4 - 2 - 0:  There are 15 ways to do this [ C(6,4) = 15 ]

3 - 3 - 0:  There are 20 ways to do this [ C(6,3) = 20 ]

4 - 1 - 1:  There are 15 ways to do this [ C(6,4) = 15 -- Once you choose the 4 balls to put into the first box, you just put one of the remaining balls into the second box and the remaining ball into the third box. This gives no more possibilities than choosing 4 balls into the first box and placing the remaining two balls into another box. ]

3 - 2 - 1:  There are 60 ways to do this [ C(6,3) x C(3,2) = 20 x 3 = 60 For each of the 20 ways that the first 3 balls can be chosen, there are 3 ways to choose 2 of the remaining 3 balls to put into the seoncd box. ]

2 - 2- 2:  There are 90 ways to do this [ C(6,2) x C(4,2) = 15 x 6 = 90 ]

1 + 6 + 15 + 20 + 15 + 60 + 90 = 207

Jun 21, 2017