There are balls of 20 colors in an urn. Each color has 5 balls so 100 balls total. You draw 10 balls without replacement from the urn. What's the probability that the 10 balls all have different colors?

I think the answer is something like $\frac{100\times (100-5)\times(100-2\times5)...\times (100-9\times5)}{100\choose 10}$, but I'm missing some multiplication factor.

Any tips appreciated. Thanks!

  • $\begingroup$ Hint: You can factor $\Pr[\textrm{all balls unique}] = \Pr[\textrm{first ball unique}] \Pr[\textrm{second ball unique} \mid \textrm{first ball unique}] \dots$ $\endgroup$
    – jII
    Commented Jul 9, 2019 at 21:43

1 Answer 1


First I'll count the right selections with ordering:

  • The first has 100 options. (all are OK)
  • The second has 95 options (1 gone, and 4 forbidden of the colour of the first)
  • The third has 90 options (1 extra gone, 4 extra forbidden ones), and this pattern continues. (so $100-5(n-1)$ options for ball $n$)

So in order we have $$100 \times 95 \times 90 \times \ldots \times 55 = 5^{10}(20 \times 19 \times 18 \times \ldots \times 11)=\frac{5^{10} 20!}{10!}$$ ordered options and we divide that number by $10!$ because we have that many ways to get the same end result given 10 different balls.

To get a probability, divide by $\binom{100}{10}$.

So $$p=\frac{5^{10}\binom{20}{10}}{\binom{100}{10}}$$

which is approximately $0.10423$.

BTW, your own first formula was correct it seems, except that the numerator is $10!$ too large: upper side is ordered options and below is unordered.

  • $\begingroup$ you made an error simplifying. 20!/10! does not equal 20 choose 10. $\endgroup$ Commented Jun 27, 2022 at 21:40

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