# Probability of selecting 7 blocks such that there is at least one from each 5 color

There are 30 blocks in a bag, which can be one of 5 colors, and are all distinct: each of the 5 colors red, yellow, green, blue, purple has 6 colored blocks numbered 1 through 6. We randomly select 7 out of the 30 blocks. If blocks are equally likely to be chosen what is the probability that we select 7 blocks such that there is at least one block for every color (red, yellow, green, blue, purple) out of the 7?

Is the number of possibilities (sample space) for 7-block choosings $\binom{30}{7}$ and therefore the probability of one particular outcome 1/$\binom{30}{7}$? Then, how do we count how many valid 7-block choosings there are to satisfy the constraint?

Hint

First figure out how many different ways you can get 7 blocks with 5 different colors without looking at the number of the block. There are two main ways of doing so:

1. Have 1 color with 3 blocks, and 4 colors with 1 block. There are of course 5 ways to do this: the color that has 3 blocks is 1 of 5

2. Have 2 colors that have 2 blocks each, and 3 colors with 1 color each ... there are $\binom{5}{2}=10$ ways this can be done: the colors with the two blocks are 2 out of 5

Then, figure out how many different numbered blocks you can get for each of the colors for these two different cases:

1. The 1 color with 3 blocks has $\binom{6}{3} = 20$ possibilities. The other 4 colors have 1 block so 6 possibilities. So that's $20*6^4$ possibilities total

2. The 2 colors with 2 blocks have $\binom{6}{2} = 15$ possibilities each, and the 3 colors with 1 block have 6 possibilities each. So that's $15^2*6^3$ possibilities total.

Since we established that there are 5 ways of getting case 1, and 10 ways of getting case 2, you get $5*20*6^4+10*15^2*6^3$ total ways of getting 7 blocks with 5 different colors.