A question regarding picking balls from an urn Five balls are randomly chosen, without replacement,
from an urn that contains 5 red, 6 white, and
7 blue balls. Find the probability that at least one
ball of each color is chosen.
I suppose I should use the Inclusion–exclusion principle but I would like to know why my initial attempt was wrong.
$$\frac{\binom{5}{1}\binom{6}{1}\binom{7}{1}\binom{15}{2}}{\binom{18}{5}}$$ Why is this wrong? Since there has to be at least 1 ball of each color we have $\binom{5}{1}\binom{6}{1}\binom{7}{1}$ ways of picking them. For  each combination there remains 2 balls to be picked from the remaining $15$, so $\binom{15}{2}$ ways. The sample space consists of $\binom{18}{5}$ outcomes. Or so I though, clearly I am wrong. What did I mess up?
 A: Your solution does not address double counting properly.
I don't know if this clarifies it: If you take one of 5 red balls, and from the last two drawings, one ball is red, those can't be distinguished.
You should indeed use inclusion exclusion on the selections where 1 or 2 colors are missing.
A: You're over-counting.
Your approach takes some red ball (from the 5 in the urn), then some white (from the 6) then some blue (from the 7).  That's fine.  Then from the 15 remaining, you choose 2.  This certainly gets you a selection of 5 from the 18 total balls. 
The problem is that you don't want to count this same selection again because that will incorrectly increase the overall probability.  But you will get it again: you choose a different first red ball, but the ${15 \choose 2}$ at the end chooses the original red ball again and you get the same set as before.
For example, suppose you select the five balls $r_1, w_1, b_1, r_2, w_2$ via your formula.  Next you select $r_2, w_1, b_1, r_1, w_2$.  But this shouldn't be counted again.
