how many ways are there to place identical balls in to urn if each urn has exactly same amount of balls?

Question

I am currently struggling with the question in combinatorics: there are 24 balls, 12 of which are red and 12 are blue, and 3 urns. How many ways are there to distribute these balls independently of their color into the three urns if each urn contains 8 balls?

My answer would be $\binom{24}{8}\cdot\binom{16}{8}\cdot\binom{8}{8}$. Is this correct?

My problem with my answer is that i cannot distuinguish the red or blue balls from each other so I think I am counting too many possibilities this way. So for example if there are 12 boys and 12 girls and we ask how many ways are there to form 3 groups of 8 people each regardless of their gender the above expression would be correct because I can distinguish between the individual students, right? I would be happy if somebody could confirm or help. Thank you

Answer 1

Your proposed answer of $\pmatrix{24 \\ 8} \pmatrix{16 \\ 8} \pmatrix{8 \\ 8} = 9,465,511,770$ is wildly wildly off.

For indistinguishable urns, there are 13 ways.

Consider just the red balls. If we know where they lie, then we know where the blue balls lie. Order the urns by the one that contains the largest number of reds, then the second largest number of reds, then the least number of reds. Actually, $i \geq j \geq k$ where $i + j + k = 12$.

Here they are:

• 840
• 831
• 822
• 750
• 741
• 732
• 660
• 651
• 642
• 633
• 552
• 543
• 444

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Here's the Mathematica code that provides these, once you realize that you must count entries with $0$.

    PadRight[#, 3] & /@ 
    Select[IntegerPartitions[12, {2, 3}], #[[1]] < 9 &]

This computes the integer partitions of 12 that are of length 2 or 3. Then it selects only those whose largest entry are less than $9$ (as only 8 can fit in an urn). The PadRight just adds $0$ to each list to make the length 3 (as there are three urns).

If you want the mathematical formula:

$$\sum\limits_{i=4}^8 \sum\limits_{j = \lceil \frac{12-i}{2} \rceil}^{\min [i, 12-i]} 1 = 13,$$

where $\lceil n\rceil$ is the "ceiling" function... the smallest integer above or equal to $n$.

Here's a plot of the discrete configurations for the first two urns, $i$, and $j$. (The third urn is simply $12-i-j$ and isn't shown.)

Answer 2

Hint. Suppose the urns are different. Let $r_1,r_2,r_3$ be the number of red balls in the first, second, third urns. (Note: if the urns were identical, then "first", "second", "third" would be meaningless.) Then $r_1,r_2,r_3$ are non-negative integers satisfying $$r_1+r_2+r_3=12\ ,\quad r_1\le8\ ,\quad r_2\le8\ ,\quad r_3\le8\ .$$ Counting the number of solutions here is a standard combinatorial technique which you may well have seen in your studies.

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So I think now you understand why, ignoring the inequalities above, the number of solutions is $C(14,2)$ (or $C(14,12)$, same thing). Now let's reintroduce the first inequality, $0\le r_1\le8$. You need to subtract those of the solutions you have already counted which violate this inequality, that is, those with $r_1\ge9$. If this is the case, then writing $s_1=r_1-9$ gives $$s_1+r_2+r_3=3\ .$$ Can you count the number of solutions here and subtract from your previous total? Can you then work out what to do about the other two inequalities?

Answer 3

First distribute the red balls across the three urns. Each urn can contain at most $8$ balls, so the number of ways to distribute the red balls is precisely the number of solutions to $$a+b+c=12,$$ in integers $0\leq a\leq b\leq c\leq8$. As each urn must contain precisely $8$ balls, the only way to then distribute the blue balls is to fill each urn up to $8$ balls.