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Andrew, a math teacher, needs to choose a team to send to a math compe

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Andrew, a math teacher, needs to choose a team to send to a math compe [#permalink]

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New post 06 Mar 2017, 01:40
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Question Stats:

71% (01:23) correct 29% (02:02) wrong based on 85 sessions

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Andrew, a math teacher, needs to choose a team to send to a math competition. \ Two of his best students, Maria and Nathan, refuse to work together. If Andrew has twelve students to choose from, and he must choose either Maria or Nathan so that he sends a strong team, how many possible five-person teams can he send?

A. 120
B. 252
C. 420
D. 792
E. 1320
Spoiler: OA

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Re: Andrew, a math teacher, needs to choose a team to send to a math compe [#permalink]

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New post 06 Mar 2017, 02:59
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Since either Maria or Nathan has to be part of the 5 member team,

Andrew can select 4 team members from the remaining 10 team members and add either Maria or Nathan as the 5th team members

To select 4 team members out of 10 = 10C4 = 10 ! / ( 6! * 4 !) = 210
Since either Maria or Nathan can be the 5th team member, total number of teams = 210 * 2 = 420

Answer is C. 420
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Re: Andrew, a math teacher, needs to choose a team to send to a math compe [#permalink]

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New post 07 Mar 2017, 17:21
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Bunuel wrote:
Andrew, a math teacher, needs to choose a team to send to a math competition. \ Two of his best students, Maria and Nathan, refuse to work together. If Andrew has twelve students to choose from, and he must choose either Maria or Nathan so that he sends a strong team, how many possible five-person teams can he send?

A. 120
B. 252
C. 420
D. 792
E. 1320



We need to determine the number of ways Andrew can select a team with either Maria or Nathan.

If Maria is selected for the team but Nathan is not, the team can be selected in 10C4 ways (since Maria definitely makes the team and Nathan does not):

10C4 = (10 x 9 x 8 x 7)/4! = (10 x 9 x 8 x 7)/(4 x 3 x 2 x 1) = 10 x 3 x 7 = 210 ways

Since there are an additional 210 ways in which the team could be created (with Nathan and without Maria), Andrew can select the team in 210 + 210 = 420 ways.

Answer: C
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Re: Andrew, a math teacher, needs to choose a team to send to a math compe [#permalink]

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New post 07 May 2018, 07:44
ScottTargetTestPrep wrote:
Bunuel wrote:
Andrew, a math teacher, needs to choose a team to send to a math competition. \ Two of his best students, Maria and Nathan, refuse to work together. If Andrew has twelve students to choose from, and he must choose either Maria or Nathan so that he sends a strong team, how many possible five-person teams can he send?

A. 120
B. 252
C. 420
D. 792
E. 1320



We need to determine the number of ways Andrew can select a team with either Maria or Nathan.

If Maria is selected for the team but Nathan is not, the team can be selected in 10C4 ways (since Maria definitely makes the team and Nathan does not):

10C4 = (10 x 9 x 8 x 7)/4! = (10 x 9 x 8 x 7)/(4 x 3 x 2 x 1) = 10 x 3 x 7 = 210 ways

Since there are an additional 210 ways in which the team could be created (with Nathan and without Maria), Andrew can select the team in 210 + 210 = 420 ways.

Answer: C


Hi,

Can you check please why I get wrong result? Here is how I calculate:
Favorable outcomes=Total outcomes-Unfavorable outcomes (when Maria and Nathan are together in the team)
Favorable outcomes=12!/7!5!-10!/3!7! (we have placed Maria and Nathan to the team of five and need to add other 3 people from the rest of 10 students) = 672

Thanks.
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Re: Andrew, a math teacher, needs to choose a team to send to a math compe   [#permalink] 07 May 2018, 07:44
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