Balls into Boxes

Probability Level 1

How many ways are there to put $10$ differently colored balls into $2$ identical boxes such that neither box is empty?

Bonus: Generalize this for ' $n$ ' balls into $2$ boxes.

The answer is 511.

2 solutions

Melissa Quail
Sep 13, 2015

There are two choices for which urn to put each of the 10 balls into so there are 1024 ways to place the ten balls if the two urns are different. However, one of these ways will involve putting all the balls in the first urn and another way will involve putting all of the balls in the other urn. These cases will leave one of the urns empty so we have to discount them leaving 1022 cases. The two urns are identical so we have counted each possibility twice and therefore the final answer is 1022/2 = 511.

What would you do if there were 3 boxes?

e r - 2 years, 11 months ago

Formula for r=3 is- S(n,3)= [(3^(n-1)+1)/2] -2^(n-1)

Akash Tyagi - 2 years, 1 month ago

45 minus 2, for the 2 cases the boxes are empty (left and right); so 43 is the answer I got. I used factorials. 10!/2!(10-2)! = 45. This is the number of unordered pairs, or subsets, of a set of 10 distinct objects. Placing n numbered balls into two urns is equivalent to the number of 2 subsets of n.

Richard Meese - 4 years, 9 months ago

Ritabrata Roy
Sep 3, 2018

At general case, if n distinguishable balls are put in 2 distinguishable bins then the number of arrangement is 2^n-2

But in question the bins are indistinguishable so the number of arrangement is half of the previous case i.e. 2^(n-1) -1

Balls into Boxes

The answer is 511.

2 solutions

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