# 6 numbered green balls, 2 indistinguishable blue balls (with restrictions)

I've been really stuck on this problem. My current thinking is to first place the blue balls, then try and distribute the green balls across the boxes without restrictions? Any help would be greatly appreciated!

Problem: 6 numbered green balls and 2 indistinguishable blue balls are to be placed in five labelled boxes.

Restrictions

• each box contains at most one blue ball
• none of the boxes are left empty

Find the number of placements?

• Just fixed it, thanks for pointing it out!
– bec7
Commented Apr 1, 2021 at 9:09
• Since only two of the five boxes will receive a blue ball, you have to take the restriction that at least one green ball must be placed in each of the other boxes into account. Commented Apr 1, 2021 at 9:15

Let it be that the boxes are labeled with $$1,2,3,4,5$$.

Then start with placing a blue ball in box $$4$$ and a blue ball in box $$5$$.

Under this condition further for $$i=1,2,3$$ let $$B_i$$ denote the set of arrangements for the green balls leaving box $$i$$ empty.

Then with inclusion/exclusion and symmetry we find that: $$5^6-|B_1\cup B_2\cup B_3|=5^6-3|B_1|+3|B_1\cap B_2|-|B_1\cap B_2\cap B_3|=$$$$5^6-3\cdot4^6+3\cdot3^6-2^6$$is the number of "green ball" arrangements that guarantee non-empty boxes.

We started with putting blue balls in boxes $$4$$ and $$5$$ but of course there are actually $$\binom52$$ possibilities for this, so the final outcome is:$$\binom52\left(5^6-3\cdot4^6+3\cdot3^6-2^6\right)=54600$$

• I don't think the question specifies that each box must contain at least one green ball Commented Apr 1, 2021 at 11:32
• @trueblueanil Indeed not. But it specifies that every box contains at least one ball. In my setup the boxes $4$ and $5$ already contain a (blue) ball on forehand, so left is that the boxes $1,2,3$ (not necessarily $4$ and $5$) must contain at least one green ball. Commented Apr 1, 2021 at 11:41
• Well, at least my clumsy attempt's answer tallies with yours ! (+1) Commented Apr 1, 2021 at 12:37
• @trueblueanil That is a relief for both of us then ;). Commented Apr 1, 2021 at 12:44

I assume all the balls need to be placed subject to the given restrictions

Case 1: $$3$$ non-empty "green boxes"

Choose $$3$$ of the $$5$$ boxes to fill with green balls (and and put the blue balls in the remaining two

$$\binom53\left[3^6 - \binom312^6 + \binom321^6\right]$$

Case 2: $$4$$ non-empty "green boxes"

In a similar manner, green balls in $$\binom54\left[4^6 - \binom413^6 + \binom422^6 - \binom431^6\right]$$

Multiply by $$\binom41$$ to account for placing a blue ball in a "green box" apart from the one in an empty box

Case 3: $$5$$ non-empty "green boxes

The blue balls van now be placed in $$\binom52$$ ways, thus

$$\binom52\left[5^6 - \binom514^6 +\binom523^6 - \binom532^6 +\binom541^6\right]$$

Adding up, we get $$\fbox{54600}$$

If we consider the blue balls distinct, total number of arrangements of $$\small 8$$ balls in $$\small 5$$ bins where no bins are empty = $$\displaystyle \small 5! \cdot {8 \brace 5} = 126000$$.

$${ \brace }$$ represents Stirling Number of the second kind (wiki).

Number of arrangements where both blue balls are together in a bin is given by $$\displaystyle \small 5! \cdot {7 \brace 5} = 16800$$ arrangements.

Number of arrangements with no bins having more than one blue ball $$\displaystyle \small = 126000 - 16800 = 109200$$

But we started by considering both blue balls distinct but they are not and hence total number of desired arrangements $$\displaystyle \small = \frac{109200}{2} = 54600$$.