20 two colored balls into 4 bins

Question

We have $16$ red identical balls and $4$ black identical balls. We have $4$ bins and put all balls randomly into the bins such that each bin contains $5$ balls at the end. Each arrangement has the same probability.

How many ways are there to distribute the balls into the bins? What is the probability that all black balls are in different bins?

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My idea was to realize that the red balls have no influence, so the number of all arrangements is simply ${4+4-1\choose 4}$. As all black balls are identical there is only one way that all black balls are in different bins, hence the probability is $\frac{1}{{7\choose 4}}$. But this Sound too easy... what do I miss? Or is it correct?

Answer 1

When we work over statistics, it does not matter whether the balls or bins are identical or not, because all of them are seen as distinct. So, the denominator part is $4^{20}$.

Now, lets think about the numerator. We have $16$ distinct red balls and $4$ distinct black balls to disperse $4$ distinct bins where all black balls are in different bins .Then we can disperse those $4$ blacks into $5$ bins by $P(4,4)$ ways. The rest reds can be dispersed bu $4^{16}$ ways. So $$\frac{24 \times 4^{16}}{4^{20}}$$

With restriction where each bins have exactly $5$ balls:

Now, if each bins contains exactly $5$ balls, then we have actually $20$ distinct balls, so we can separate them into $4$ bins where each have exactly $5$ balls: $$\binom{20}{5}\binom{15}{5}\binom{10}{5}\binom{5}{5}$$

Now, lets disperse those blacks into those $4$ bins by $P(4,4)$.After that, diperse those $16$ reds into $4$ bins where each bins have $4$ reds. So, the numerator is $$P(4,4)\times \binom{16}{4}\binom{12}{4}\binom{8}{4}\binom{4}{4}$$

The answer is $$\frac{P(4,4)\times \binom{16}{4}\binom{12}{4}\binom{8}{4}\binom{4}{4}}{\binom{20}{5}\binom{15}{5}\binom{10}{5}\binom{5}{5}}=0.128999..$$

Answer 2

As far as I am concerned, I cannot agree with the previous answer.

For convenience, let's name the four bins A, B, C, D. The first approach supposes that each bin is identical (4 balls in A are considered the same as 4 balls in D, this is ridiculous), we can quickly know that $C_{4+4-1}^{4}$ is not a great solution for the following two reasons:

1. $3$ black balls in A and $1$ in B is not considered the same distribution as $1$ in A and $3$ in B.
2. $4$ black balls in A is not considered the same as $4$ black balls in D.

Thus, we can just simply enumerate the distributions, namely $$ 4, 3+1, 2+2, 2+1+1, 1+1+1+1,$$ So there are just 5 possibilities. Which is ridiculous.

The second explanation is that the bins are different. Well, in this case, you are absolutely right. I can enumerate all possibilities to prove it:

1. $4$ balls all in the same bin, $4$ distributions.
2. $3+1$, $A_{4}^{2}=12$ distributions.
3. $2+2$, $C_{4}^{2}=6$ distributions.
4. $2+1+1,$ $C_{4}^{3}*C_{3}^{1}=12$ distributions.
5. $1+1+1+1$ 1 possibility.

All in all, $4+12+6+12+1=35=C_{7}^{4}$.

Answer 3

I consider as distinct the following cases:

• all 4 black balls in the same bin : 4 cases

• 3 black balls in the same bin : $4\times 3 = 12$ cases

• 2 black balls in one bin, 2 black balls in the other bin : $\binom{4}{2} = 6$ cases

• 2 black balls in the same bin, and other 2 black balls in different bins: $4\times 3 = 12$ cases (choose a bin for 2 black balls, then choose a bin without black balls)

• all black balls in 4 different bins: 1 case

Total number of cases : $4+ 12 +6+12+1=35$

Probability: $P=\frac{1}{35}$

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PS: I am not happy with my solution. It assumes that all distinct cases are equally likely, which is hardly true in a real world. I just followed the problem statement. Maybe the problem statement is intentionally confusing.

Answer 4

A very useful ploy is to fit a given problem into a well recognised framework.

Here we can consider it as distributing $16$ red and $4$ black cards into $4$ hands, and the answers easily flow out

How many ways are there to distribute the balls into the bins?

$$\binom{20}5\binom{15}5\binom{10}5\binom55 $$

What is the probability that all black balls are in different bins?

$$\frac{\binom{16}4\binom41\;\binom{12}4\binom31\;\binom84\binom21\;\binom44\binom11}{\binom{20}5\binom{15}5\binom{10}5\binom55 }=\frac{125}{969}$$