drawing chips from a bag without replacement problem: A bag contains 7 black, 8 orange, and 9 red chips. You draw five chips from the bag at random and without replacement. What is the probability that you draw at least one chip of each color?
I am new to probabilities, and am finding myself kind of lost on this question. From what I have been able to gather from researching similar problems I've been led to believe that a solution might be (using combinatorics):
$\dfrac{{{7}\choose{1}} {{8}\choose{1}} {{9}\choose{1}}{{24}\choose 2}}{{{24}\choose 5}}$
But I'm not sure if it's correct or how this method works. This is my first time encountering a problem like this and I would really appreciate any help I could get. Thank you.
 A: An easier approach might be to use inclusion exclusion.  What is the probability of not getting any black?  $$\frac{{7 \choose 0}\cdot {17 \choose 5}}{24 \choose 5}$$
Then the probability of not getting orange is similar, and likewise red.
If you add those up, you have triple counted the cases where only $1$ color was picked, and so you should subtract the extras.  Then $1-$ that is your answer.
A: The answer given by the original poster is wrong for two reasons:
(1) After selecting one from each color, you would be left with only $(21)$ chips,
so the 4th factor in the numerator should be $\binom{21}{2}.$
(2) Even with the above correction, the answer is still wrong, because of
over-counting.  That is, suppose that you label the black chips
$b_1, b_2, \cdots.$  The time when the first black chip selected is $b_1$ and
the 2nd black chip (from the remaining 21) selected is $b_2$ will also be
inaccurately counted separately when the 1st black chip selected is $b_2$
and the 2nd black (from the remaining 21) selected is $b_1$.
To avoid this type of pitfall, it is better to take the approach outlined in
the answer of Robert The Tutor.
