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
Brent Hanneson - Creator of GMATPrepNow.com
