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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
I'm no expert, just trying to work on my skills. If I've made any mistakes please bear with me.
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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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