Problem Solving

Hi swerve,

We're told that a certain team consists of 4 professors and 6 teaching assistants. We're asked for the number of 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.). There are a couple of different ways to approach this prompt - the fastest would likely be to determine all of the possible groups of 3 (without any restrictions) and then subtract the ones that don't fit the specific restrictions that the prompt gives us.

Since we're forming groups of 3 from 10 total people, there are 10c3 = 10!/3!(10-3)! = (10)(9)(8)/(3)(2)(1) = 120 possible groups of 3. However, this total includes groups that consist of only 3 teaching assistants.

With 6 teaching assistants, there would be 6c3 = 6!/3!(6-3)! = (6)(5)(4)/(3)(2)(1) = 20 groups of 3 teaching assistants. These groups don't include at least one professor, so they must be removed from the total.

Total = 120 - 20 = 100 possible groups that fit the given restrictions.

Final Answer: B

GMAT assassins aren't born, they're made,
Rich

swerve 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 ways)

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 ways)

ASIDE: If anyone is interested, we have a free video on calculating combinations (like 6C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

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