# Probability of choosing different colors when there are more colors than balls per color

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!

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

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.