A bag contains 100 marbles, and each marble is one of three different colors. If you were to draw three marbles at random, the probability that you would get one of each color is exactly 20 percent.
How many marbles of each color are in the bag?
Let's say that the number of each color is $A,$ $B$ and $C,$ with $A + B + C = 100.$ The number combinations of one marble of each color would thus be $ABC.$ The total number of possibilities for pulling three marbles out is $\binom{100}{3}.$ So if the probability is exactly $20\%$ then we would have $$\frac{ABC}{\binom{100}{3}} = 20\%$$ that is $$ABC = 0.2 \cdot \frac{100 \cdot 99 \cdot 98}{6} = 32,340.$$
Even after decomposing into prime factors, there are a lot of ways to possible come up with $A$, $B$ and $C$ such that $ABC = 32,340 = 2^2 \cdot 3 \cdot 5 \cdot 7^2 \cdot 11.$ However, there are not so many possible combinations that it makes sense to do something more rigorous than the all-important "guess and check" method, which eventually ends you up with $A = 21,$ $B = 35,$ and $C = 44.$
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