Probability of drawing at least 1 red, 1 blue, 1 green, 1 white, 1 black, and 1 grey when drawing 8 balls from a pool of 30? Given a pool of 30 balls (5 of each color). When drawing 8 balls without replacement, what is the probability of getting at least one of each color?
Related: Probability of drawing at least one red and at least one green ball.
When drawing more than 2 colors you need to exclude overlapping 'hands'. Thus when finding the probability of drawing no red, you can have a hand made up of blue, green, white, black and grey. But when you are determining the probability of drawing no blue you draw from red, green, white, black, grey. So you need to exclude all green, white, black, grey hands as they have already been counted. And the same for the other colors as well.
The other complexity of the problem is that since there are only 5 of each color, no draw will only include balls of the same color.

The quickest approach is to count ways to get $6$ colors on $8$ balls. There are essentially two cases, $\langle 3,1,1,1,1,1\rangle$ and $\langle 2,2,1,1,1,1\rangle$. There are $\binom{6}{1}\binom{5}{3}\binom{5}{1}^5=187,500$ ways to get the first  case. The $\binom{6}{1}$ counts the number of ways of choosing one color to get three balls, and the rest is the number of ways of choosing three balls from that one color and one for each of the other colors.
The second case has $\binom6 2\binom 5 2^2\binom 5 1^4=937,500$ different ways. There are $\binom 6 2$ ways to choose two colors to receive two balls, and the rest is the number of ways of choosing two balls from each of those colors and one ball from the others.
So the total cases are $1,125,000$, out of $\binom{30}{8}=5,852,925$. That gives a probability of about $0.1922$.
