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Mrs. Pearson has 4 boys and 5 girls in her class.

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jedit
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Mrs. Pearson has 4 boys and 5 girls in her class. [#permalink] New post   19 Jul 2017, 08:49
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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
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C
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Re: Mrs. Pearson has 4 boys and 5 girls in her class. [#permalink] New post   19 Jul 2017, 09:10
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jedit wrote:
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


Hi,

I really dont know the purpose of" 1 boy n 1 girl leave"?
Does it make Q complicated- NO.
Hard-NO...
Tests something extra- NO... Just subtracting 1 from 4 n 5..

So there are 3 boys and 4girls left.... Choose 2 from each.
3C2*4C2=3*6=18

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Thanos7
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Re: Mrs. Pearson has 4 boys and 5 girls in her class. [#permalink] New post   20 Jul 2017, 00:23
Should be C(18),
Picking 2 boys from 3 boys=3C2
Picking 2 girls from 4 girls=4C2
Total arrangements=3C2*4C2=3*6= 18

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Re: Mrs. Pearson has 4 boys and 5 girls in her class. [#permalink] New post   25 Jul 2017, 11:52
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jedit wrote:
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


Since one girl and one boy leave before she can make a selection, she now has 4 girls and 3 boys left. Thus, we need to determine the number of ways to select 2 girls from 4 and 2 boys from 3.

The girls can be selected in 4C2 = (4 x 3)/2! = 6 ways and the boys can be selected in 3C2 = 3 ways. Thus, the total number of ways to select the committee is 6 x 3 = 18 ways.

Answer: C
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Re: Mrs. Pearson has 4 boys and 5 girls in her class. [#permalink] New post   02 Jul 2020, 22:48
originally there were 4 boys and 5 girls.
1 of each leave, so now there are 3 boys and 4 girls.
Need to choose 2 boys out of 3 = \(3C2 = 3\)

Need to choose 2 girls out of 4 = \(4C2 = 6\)

Total combinations = \(3 \times 6 = 18\)

OA, C
jedit wrote:
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
GMAT Club Bot
gmatclubot
Re: Mrs. Pearson has 4 boys and 5 girls in her class. [#permalink]
02 Jul 2020, 22:48
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