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Bunuel
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Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
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4C2 is not equal to 18
Elihof
1 girl and 1 boy leave so you are left with 3 boys and 4 girls:

- Number of unique combinations of 2 out of 3 boys = 3C2 = \( \frac{ 3!}{2! } \) \(= 6\)

- Number of unique combinations of 2 out of 4 girls = 4C2 = \( \frac{ 4!}{(2!*2!) } \) \( = 18\)

Total options = girl combo x boy combo = \(3 * 6 = 18\)

IMO: C
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Thanks! Mistake in typing it down, adjusted it :)
hari0616
4C2 is not equal to 18
Elihof
1 girl and 1 boy leave so you are left with 3 boys and 4 girls:

- Number of unique combinations of 2 out of 3 boys = 3C2 = \( \frac{ 3!}{2! } \) \(= 6\)

- Number of unique combinations of 2 out of 4 girls = 4C2 = \( \frac{ 4!}{(2!*2!) } \) \( = 18\)

Total options = girl combo x boy combo = \(3 * 6 = 18\)

IMO: C
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Bunuel
Mrs. Pearson has 4 boys and 5 girls in her class. She is to choose 2 boys and 2 girls to serve on her grading committee. If one girl and one boy leave before she can make a selection, then how many unique committees can result from the information above?

A. 9
B. 12
C. 18
D. 22
E. 120


­

Basically Mrs Pearson has to choose 2 boys from 3 boys and 2 girls from 4 girls
Number of ways = 3C2 * 4C2 = 3 * 6 = 18 ways
Ans C
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