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A certain team consists of 4 professors and 6 teaching assistants. How

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A certain team consists of 4 professors and 6 teaching assistants. How  [#permalink]

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New post 25 May 2016, 04:22
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A certain team consists of 4 professors and 6 teaching assistants. How many different teams of 3 can be formed in which at least one member of the group is a professor? (Two groups are considered different if at least one group member is different.)

A. 48
B. 100
C. 120
D. 288
E. 600

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Re: A certain team consists of 4 professors and 6 teaching assistants. How  [#permalink]

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New post 25 May 2016, 05:36
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The groups can be formed as
PTT + PPT + PPP where P=professor and T=Teaching assistant
Number of different teams of 3 can be formed in which at least one member of the group is a professor
= 4C1* 6C2 + 4C2*6C1 + 4C3
= 4!/3! * 6*5/2 + 4*3/2 * 6 + 4
=60 +36 + 4
=100

Answer B
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Re: A certain team consists of 4 professors and 6 teaching assistants. How  [#permalink]

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New post 26 May 2016, 01:58
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Atleast one professor = Total - No professor

i.e, 10C3 - 6C3 = 120 -20 =100

ans : B
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Re: A certain team consists of 4 professors and 6 teaching assistants. How  [#permalink]

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New post 26 May 2016, 09:14
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Bunuel wrote:
A certain team consists of 4 professors and 6 teaching assistants. How many different teams of 3 can be formed in which at least one member of the group is a professor? (Two groups are considered different if at least one group member is different.)

A. 48
B. 100
C. 120
D. 288
E. 600


At least one professor= Total possibility- All teaching assistants

Total possibility= 10C3= 120
Teaching assistant= 6C3= 20

AT least one professor= 120-20= 100

B is the answer
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Re: A certain team consists of 4 professors and 6 teaching assistants. How  [#permalink]

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New post 26 May 2016, 10:03
Bunuel wrote:
A certain team consists of 4 professors and 6 teaching assistants. How many different teams of 3 can be formed in which at least one member of the group is a professor? (Two groups are considered different if at least one group member is different.)

A. 48
B. 100
C. 120
D. 288
E. 600


Number of different teams of 3 that can be formed in which at least one member of the group is a professor = Total Number of different teams of 3 that can be formed - Number of different teams of 3 that can be formed in which no member of the group is a professor
Total Number of different teams of 3 that can be formed = 10!/7!3! = 120
Number of different teams of 3 that can be formed in which no member of the group is a professor = 6!/3!3! = 20
Number of different teams of 3 that can be formed in which at least one member of the group is a professor = 120 - 20 = 100
Answer - B
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A certain team consists of 4 professors and 6 teaching assistants. How  [#permalink]

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New post 04 Jun 2018, 10:47
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Bunuel wrote:
A certain team consists of 4 professors and 6 teaching assistants. How many different teams of 3 can be formed in which at least one member of the group is a professor? (Two groups are considered different if at least one group member is different.)

A. 48
B. 100
C. 120
D. 288
E. 600


When we see a counting question involving "at least", we should consider using the nice rule:
Total number of outcomes that FOLLOW a rule = (TOTAL number of outcomes that IGNORE the rule) - (number of outcomes that BREAK the rule)

Here, we have: Total number of teams with AT LEAST one professor = (TOTAL number of teams with ANY NUMBER of professors) - (number of teams with ZERO professors)

TOTAL number of teams with ANY NUMBER of professors
W'ere ignoring the rule that talks about the number of professors on a team.
There are 10 people in total, and we must select 3 to be on a team.
Since the order in which we select the people does not matter, we can use COMBINATIONS
We can select 3 people from 10 people in 10C3 ways (= 120 teams)

Number of teams with ZERO professors
This means all 3 team members must be assistants
There are 6 assistants in total, and we must select 3 of them to be on a team.
Since the order in which we select the assistants does not matter, we can use COMBINATIONS
We can select 3 assistants from 6 assistants in 6C3 ways (= 20 teams)

ASIDE: If anyone is interested, we have a free video (below) on calculating combinations (like 6C3) in your head

Total number of teams with AT LEAST one professor = (TOTAL number of teams with ANY NUMBER of professors) - (number of teams with ZERO professors)
= 120 - 20
= 100

Answer: B

Cheers,
Brent

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