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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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70% (01:35) correct 30% (02:04) wrong based on 141 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

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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]

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New post 24 Sep 2018, 22:15
Before you even get to the restriction on this problem, first identify that it is a combination problem - the order does not matter, but rather it's just about who's on the team. So prepare to use the Combinations formula: N!/K!(N−K!)

Whenever you encounter a combinations problem with restrictions, you have two major options:

Quote:
1) Solve for the number of unrestricted combinations, then find and subtract the number of restricted combinations. Here the restricted groups are "neither Nathan nor Maria is on the team" and "both Nathan and Maria are on the same team."


Quote:
2) Solve directly for the combinations that do not violate the restriction. Here that means considering two case: Maria is on the team and Nathan is not, or Nathan is on the team and Maria is not.

For this problem, note that the calculation will be the same for both options in #2: in each case, one spot is already "spoken for," one person is no longer eligible to be part of the pool of N candidates, and 10 people are left for the 4 remaining spots. Therefore it's best on this problem to proceed by solving directly for the combinations that do not violate the restriction.


That calculation will look like:

Maria is on the team, Nathan is not: 10 candidates for 4 remaining spots = 10!/4!6!=(10×9×8×7)/(4×3×2×1)

Nathan is on the team, Maria is not: the math is the same, with 10 candidates for 4 remaining spots. This is also 210

Therefore the correct answer is the sum of those two totals, 210+210=420

Choice C is correct.
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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 27 Sep 2018, 21:02
Hero8888 wrote:
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.


Your Favorable outcomes also include the scenario when Nathan & Maria both are not present. So you need to subtract that as well from Total outcomes.
Favorable outcomes=12!/7!5!-10!/3!7!-10!/5!5!(when both Nathan & Maria aren't present)=420
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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 29 Sep 2018, 05:54
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


METHOD-1

Total Team of 5 out of 12 candidates = 12C5 = 792
Total Team of 5 out of 12 candidates without Maria and Nathan = 10C5 = 252
Total Team of 5 out of 12 candidates with Maria and Nathan both = 10C3 = 120

Favourable Outcomes = 792 - 252-120 = 420

Answer: Option C

METHOD-2

Total Team of 5 out of 12 candidates with Maria and without Nathan = 10C4 = 210
Total Team of 5 out of 12 candidates with Nathan and without Maria = 10C4 = 210

Favourable Outcomes = 210+210 = 420

Answer: Option C
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