Overcounting in combinatorics problem Disclaimer: this is a homework exercise, please don't give me the full solution! I'd just like a hint in the right direction.
A box contains 30 red balls, 30 white balls, and 30 blue
balls. If 10 balls are selected at random, without replacement,
what is the probability that at least one color will be
missing from the selection?

My reasoning so far has been to look at the event that every colour is present, and then just do $1 - $ that value. We can think of it as selecting 3 balls, one from each colour, and then filling the remaining 7 spaces with any other balls. Selecting 1 ball from 30 three times is $30^3$ combinations. Then, selecting 7 balls from 87 is $C_7^{87}$ combinations, and the total combinations is their product. However, this turns out to be larger than $C_{10}^{90}$, which is the total amount of ways 10 balls can be selected from 90 balls.
Clearly, I've overcounted somewhere, but I have no idea where! I've been stuck on this for hours and I can't see where I'm going wrong.
 A: Let's try a similar problem with smaller numbers; then we can actually list all the possibilities and see which ones we've counted twice.
Suppose there are $3$ balls: 2 red balls and one blue ball. How many ways can we draw two balls (without replacement) so that at least one is red? Call the balls $r_1,r_2,$ and $b.$
Correct solution: There are three ways to draw two balls: $r_1r_2,r_1b,r_2b$, and every one of them has at least one red ball, so the answer is $3.$
Incorrect solution: We need at least one red ball, so start by drawing a red ball: $2$ possibilities. Then draw a second ball, we don't care what color it is: $2$ possibilities. So the answer is $2\times2=4.$ What went wrong? The $4$ combinations we counted are $r_1r_2,r_1b,r_2r_1,r_2b.$ Oops, the case "draw both red balls" was counted twice!
Your mistake was where you said "We can think of it as selecting 3 balls, one from each colour, and then filling the remaining 7 spaces with any other balls."
A: hint: are the balls of each color distinguishable ? 
A: The total number of ways you can pick $10$ balls out of $90$ is $\binom{90}{10}$. In how many ways can you pick $10$ balls excluding one colour and two colours, respectively?
