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It seems like we can assume that there are at least three of each color, so I'll make that assumption. I'll also make the reasonable assumtion that order doesn't matter, so black red blue (Brb) will be considered identical to black blue red (BbR).
There are 4 ways to package three pens of the same color, since there are four colors. We can label them BBB, RRR, bbb, GGG.
There are 4 ways to package three pens, all of which are different colors. This is probably easiest to see if you think of the process as choosing which color not to include, since there are four ways to do this. We can label these BRb, BRG, BbG, RbG.
The last case is a little harder, but we can see that there are 12 ways of packaging exactly two pens of the same color and a third of a different color. One way to see this is to just list them: BBR, BBb, BBG, RRB, RRb, RRG, bbB, bbR, bbG, GGB, GGR, GGb.
Another way to see this last case is to note that we first have the choice of two colors out of the four (this can be done in 4 choose 2=6 ways) and then we have the choice of which of the two colors is repeated (so two choices). By the multiplication principle, we get 12 ways total.
This exhausts all the cases, and we have 4+4+12=20 ways to package them.
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