Number of ways $6$ distinct balls can be placed in $3$ different urns such that each urn contains at least one ball.

Question

Number of ways $6$ distinct balls can be placed in $3$ different urns such that each urn contains at least one ball.

According to me,

Total number of combination possible is $3^6$

Number of ways all urns contain $0$ balls(complement of each urn gets at least one ball) is $1$ way.

So I think answer should be $3^6 - 1$

Where am I going wrong ?

There are more ways for one urn to be empty, and of course it's impossible for all urns to be empty. — Joffan, Oct 29 '17 at 21:19
this question is a duplicate, and is itself closed as a duplicate. You can find an extensive discussion of the issue starting there. — lulu, Oct 29 '17 at 21:28

score 0 · Answer 1 · answered Oct 29 '17 at 21:21

0

'All urns do not contain at least 1 ball' $\not = $ 'Not all urns contain at least 1 ball'

answered Oct 29 '17 at 21:21

Bram28

100,612
6
70
118

score 0 · Accepted Answer · answered Oct 29 '17 at 21:22

0

You need an inclusion-exclusion approach.

Suppose some given urn is empty. Then the ways that can happen are the same as the ways that $6$ distinct balls can be distributed between $2$ urns.

And suppose $2$ urns are empty. Then there is only one way that the balls can be "distributed" into the since remaining urn.

Perhaps you can see a way forward from here.

answered Oct 29 '17 at 21:22

Joffan

39,627

Why the complement trick doesn't work here. – Sagar P Oct 29 '17 at 21:23
Because it is not true that the one urn empty is the same as all urns empty. There are many permutations that leave an urn empty, if you consider my answer here. And this is a variation of a complement trick - it's still counting the ways that the constraint gets broken, so you will still need your $3^6$ calculation to kick it off. – Joffan Oct 29 '17 at 21:26

score 0 · Answer 3 · answered Oct 29 '17 at 21:38

A approach can be to first take 3 balls from this 6 balls and put them one in each urn, so we will guarany that each urn contains one ball. After this we have 3 remainings balls, where each of them have 4 possibilities : stay in first urn, in second urn, in third urn or in non urn. Daniel

Number of ways $6$ distinct balls can be placed in $3$ different urns such that each urn contains at least one ball.

3 Answers3