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

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Mrs. Pearson has 4 boys and 5 girls in her class. She is to [#permalink] New post 04 Oct 2017, 20:32
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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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Re: Mrs. Pearson has 4 boys and 5 girls in her class. She is to [#permalink] New post 04 Oct 2017, 22:05
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Bunuel 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

Kudos for correct solution.


Since one boy and one girl leaves before selection
therefore we are left with 3 boys and 4 girls out of which She has to choose two each from boy and girl

Therefore the total number of ways to select the committee = 3C2 * 4C2 = 18 ways
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Re: Mrs. Pearson has 4 boys and 5 girls in her class. She is to [#permalink] New post 19 Dec 2017, 06:37
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Bunuel 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. She is to   [#permalink] 19 Dec 2017, 06:37
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