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by hemant_rajput » Thu Jul 04, 2013 9:56 am
A entrepreneurship competition requires registering teams to have 3 team members, at least one of which must be a technology co-founder. If all team members must come from the auditorium during the meet and greet event which has 4 technologists and 9 businessmen, how many possible team submissions are possible?

(A)76
(B)100
(C)162
(D)198
(E)202
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by Brent@GMATPrepNow » Thu Jul 04, 2013 11:21 am
hemant_rajput wrote:A entrepreneurship competition requires registering teams to have 3 team members, at least one of which must be a technology co-founder. If all team members must come from the auditorium during the meet and greet event which has 4 technologists and 9 businessmen, how many possible team submissions are possible?

(A)76
(B)100
(C)162
(D)198
(E)202
NOTE: I'm assuming that every technologist is a "technology co-founder"

Let's first ignore the rule about having at least 1 technologist.
There are 13 people altogether, and we must select 3.
Since the order of the selected people does not matter, we can use combinations.
We can select 3 people from 13 people in 13C3 ways (= 286 ways)

Of course, among those 286 teams, we have included teams that do not have any technologists. So we must eliminate those teams from our total.

# of teams with zero technologists
Take the 9 businessmen and select 3 (thus creating a team with zero technologists)
We can select 3 businessmen from 9 businessmen in 9C3 ways (= 84 ways)

So, the total number of teams = 286 - 84 = [spoiler]202 = E[/spoiler]

Cheers,
Brent

Aside: If anyone is interested, we have a free video on calculating combinations (like 9C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789
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by GMATGuruNY » Thu Jul 04, 2013 11:29 am
hemant_rajput wrote:A club composed of 4 technologists and 9 businessmen plans to send a 3-member team to an entrepreneurship convention. At least one of the team members must be a technologist. How many different teams could the club send?

(A)76
(B)100
(C)162
(D)198
(E)202
I've reworded the text to reflect the intent of the problem.

Good teams = total possible teams - bad teams.

Total possible teams:
Number of options for the first member selected = 13. (Any of the 13 members.)
Number of options for the second member selected = 12. (Any of the 12 remaining members.)
Number of options for the third member selected = 11. (Any of the 11 remaining members.)
To combine these options, we multiply:
13*12*11.
Since the ORDER of the selections DOESN'T MATTER -- ABC is the same team as BCA -- we DIVIDE by the number of ways to ARRANGE the 3 members selected (3!):
(13*12*11)/(3*2*1) = 286.

Bad teams:
A bad team does NOT include at least one technologist.
In other words, a bad team is composed SOLELY OF BUSINESSMEN.
Number of options for the first businessman selected = 9. (Any of the 9 businessmen.)
Number of options for the second businessman selected = 8. (Any of the 8 remaining businessmen.)
Number of options for the third businessman selected = 7. (Any of the 7 remaining businessmen.)
To combine these options, we multiply:
9*8*7.
Since the ORDER of the selections DOESN'T MATTER -- ABC is the same team as BCA -- we DIVIDE by the number of ways to ARRANGE the 3 businessmen selected (3!):
(9*8*7)/(3*2*1) = 84.

Good teams = 286 - 84 = 202.

The correct answer is E.
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