<s>[INST] Probability with pins in a box The question is as follows: 

There are 7 red pins, 7 black pins, and 7 green pins in a box.  Pick 5 at random without replacement.  What is the probability that at least one color was not picked?

My thinking is that this could happen in two ways.  Either the 5 pins chosen were all of the same color, or the 5 pins chosen were two different colors.  In each case, at least one color is excluded. 
Clearly, there are $\binom{21}{5}$ ways of choosing the 5 pins from the box.  
In the first case, we choose the pins so that they all have the same color.  This equates to $$\binom{7}{5} + \binom{7}{5} + \binom{7}{5}= 3\binom{7}{5}.$$
In the second case, the 5 pins are chosen from two colors, excluding one.  So, we have $14$ pins to choose from and we choose $5.$  We can either pick from the red and black, the black and green, or the red and green.  This then equates to 
$$\binom{14}{5} + \binom{14}{5} + \binom{14}{5} = 3\binom{14}{5}.$$
The final answer would then be 
$$\frac{3\binom{7}{5}+3\binom{14}{5}}{\binom{21}{5}} \approx 2982.$$
Would this be the correct answer and the correct approach? Thanks in advance!  
 [/INST] There is no need to split into two cases.
Ban a color. There are 3 ways to do this.
Now choose 5 pens from the rest 14 pens. There are $\binom{14}5$ ways to do that.
However we must be aware of double counting. If we chose 5 pens from the same color, then it's counted twice. Therefore we must subtract $\binom{7}{5}\times3$.
Therefore the probability is $\frac{3\times(\binom{14}{5}-\binom75)}{\binom{21}5}$
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