Stringworm, your list of possibilities is complete--it just doesn't add up to 14 because it's the answer to a different question. You have shown all the ways to put 2 indistinguishable boys and 2 indistinguishable girls in order. What we want for our denominator is a list of all the ways we can choose 4 particular (distinguishable) children out of 3 boys and 3 girls. We aren't concerned with what order the children appear in, but we

are concerned with which children are selected. The list looks like this:

(g1)(g2)(b1)(b2)

(g1)(g2)(b1)(b3)

(g1)(g2)(b2)(b3)

(g1)(g3)(b1)(b2)

(g1)(g3)(b1)(b3)

(g1)(g3)(b2)(b3)

(g2)(g3)(b1)(b2)

(g2)(g3)(b1)(b3)

(g2)(g3)(b2)(b3)

(b1)(all 3 girls)

(b2)(all 3 girls)

(b3)(all 3 girls)

(g1)(all 3 boys)

(g2)(all 3 boys)

(g3)(all 3 boys)

Note that the first 9 choices are the ones we want in the numerator. Also note that we would

not want to spend time doing this on the test. If you can't find a faster way to do it, drop the problem!

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Dmitry Farber | Manhattan GMAT Instructor | New York

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