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!
Best Answer
First I'll count the right selections with ordering:
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.