# In how many ways can we place $10$ identical red balls and $10$ identical blue balls into $4$ distinct urns if there are no constraints?

In how many ways can we place $$10$$ identical red balls and $$10$$ identical blue balls into $$4$$ distinct urns if there are no constraints?

I have tried $$10$$ red and $$10$$ blue balls.

Total balls: $$20$$

Now, let us arrange them in a straight line.

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

We have all the balls lined up.

Now we have to divide them into $$4$$ sections.

0 0 0| 0 0 0|0 0 0 0 0 0|0 0 0 0 0 0 0 0|

(here | represents a cut which divides the balls)

This is one of the cases. We have to count all the cases.

We have a total of $$24$$ objects.

The number of way of placing $$24$$ objects are $$24!$$, but we have $$10$$ red identical balls as well as $$10$$ blue identical balls and $$4$$ identical ( | ).

So the total number of combinations are: $$\frac{24!}{10!10!4!}$$ My answer comes to $$1963217256$$, but this is wrong. I don't know where I'm wrong.

• Can you compute the number of ways to distribute the $10$ red balls among the $4$ urns? (see Stars and Bars). Compute the number of ways to distribute the $10$ blue balls among the $4$ urns in the same way. The number of arrangements we are looking for is the product. – robjohn Jul 24 at 7:07
• Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. – N. F. Taussig Jul 24 at 8:44

Two hints:

• You should consider the red balls and blue balls separately

• If using a stars and bars method, the bars are the divisions between the boxes so $$1$$ less than the number of boxes

What error did you make?

The order of the balls within each urn does not matter. However, in your attempt, you took the order of the balls in each urn into account, so your result is far too large.

Let's illustrate with smaller numbers. Suppose we have three blue and three red balls to be placed in two distinguishable urns. There are $$16$$ possibilities:

$$\begin{array}{l l} \text{urn 1} & \text{urn 2}\\ \hline bbbrrr & \\ bbbrr & r\\ bbbr & rr\\ bbb & rrr\\ bbrrr & b\\ bbrr & br\\ bbr & brr\\ bb & brrr\\ brrr & bb\\ brr & bbr\\ br & bbrr\\ b & bbrrr\\ & bbbrrr\\ r & bbbrr\\ rr & bbbr\\ rrr & bbb\\ \end{array}$$

Your approach would yield $$\frac{7!}{3!3!1!} = 144$$ since we would need one divider.

There is only way to place all six balls in the same urn. However, in your attempt, you treat $$bbbrrr$$ differently than $$brbrbr$$. Since there are $$\binom{6}{3} = 20$$ ways of arranging three blue and three red balls in the urn, you count this case twenty times rather than once.

Why are there actually $$16$$ cases?

There are four ways to distribute the blue balls since we must place $$0$$, $$1$$, $$2$$, or $$3$$ of them in the first urn and the rest in the second urn. By symmetry, there are also four ways to distribute the red balls. Hence, there are $$4 \cdot 4 = 16$$ possible distributions of three blue and three red balls in two distinguishable urns.

How should you solve the problem?

As robjohn said in the comments, distribute the blue balls and red balls separately. Let $$b_i$$, $$1 \leq i \leq 4$$, denote the number of blue balls placed in the $$i$$th urn. Then the number of ways of distributing $$10$$ indistinguishable blue balls to four distinguishable urns is the number of solutions of the equation $$b_1 + b_2 + b_3 + b_4 = 10$$ in the nonnegative integers. By symmetry, the $$10$$ indistinguishable red balls can be distributed to four distinguishable urns in the same number of ways. Since these distributions are independent (assuming each urn can hold at least $$20$$ balls), the answer is found by multiplying the number of ways of distributing the blue balls by the number of ways of distributing the red balls.

The number of ways of distributing ten indistinguishable blue balls to four distinguishable urns can be found as follows: A particular solution of the equation $$b_1 + b_2 + b_3 + b_4 = 10$$ in the nonnegative integers corresponds to the number of ways of placing $$4 - 1 = 3$$ addition signs in a row of $$10$$ ones. For instance, $$1 1 1 1 + 1 1 1 + 1 1 + 1$$ corresponds to placing $$4$$ blue balls in the first urn, three blue balls in the second urn, $$2$$ blue balls in the third urn, and one blue ball in the fourth urn, while $$+ 1 1 1 1 1 + + 1 1 1 1 1$$ corresponds to placing no blue balls in the first or third urns and five blue balls apiece in the second and fourth urns. Therefore, the number of solutions of the equation is $$\binom{10 + 4 - 1}{4 - 1} = \binom{13}{3}$$ since we must choose which three of the thirteen positions required for ten ones and three addition signs will be filled with addition signs. To finish the problem, multiply by the number of ways of distributing the red balls.