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# An entrepreneurship competition requires registering teams to have 3

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Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132619 [0], given: 12326

An entrepreneurship competition requires registering teams to have 3 [#permalink]

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26 May 2016, 12:33
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Difficulty:

65% (hard)

Question Stats:

69% (01:30) correct 31% (02:25) wrong based on 58 sessions

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An 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
[Reveal] Spoiler: OA

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Kudos [?]: 132619 [0], given: 12326

Senior Manager
Joined: 12 Mar 2013
Posts: 258

Kudos [?]: 166 [1], given: 350

Re: An entrepreneurship competition requires registering teams to have 3 [#permalink]

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26 May 2016, 14:27
1
KUDOS
Bunuel wrote:
An 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

P(at least one Tech)= p(total) - P( no tech)

=13c3 - 9c3

=198 (ans D )
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Kudos [?]: 166 [1], given: 350

Manager
Joined: 29 Nov 2011
Posts: 114

Kudos [?]: 23 [0], given: 367

An entrepreneurship competition requires registering teams to have 3 [#permalink]

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28 May 2016, 06:55
nahid78 wrote:
Bunuel wrote:
An 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

P(at least one Tech)= p(total) - P( no tech)

=13c3 - 9c3

=198 (ans D )

calculation seems to be wrong .. its 202 i guess

i put it differently
4C1 *9C2 + 4C2*9C1 + 4C3 = 202

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Manager
Joined: 18 May 2016
Posts: 67

Kudos [?]: 63 [1], given: 105

Concentration: Finance, International Business
GMAT 1: 720 Q49 V39
GPA: 3.7
WE: Analyst (Investment Banking)
An entrepreneurship competition requires registering teams to have 3 [#permalink]

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28 May 2016, 07:24
1
KUDOS
We have 3 scenarios here:

1) 1 tech & 2 businessmen: 4C1 x 9C2 = 144
2) 2 tech & 1 businessman: 4C2 x 9C1 = 54
3) 3 tech & 0 businessmen: 4C3 = 4

Total: 144+54+4 = 202

Are there any shortcuts to avoid doing all the scenarios separately?
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Kudos [?]: 63 [1], given: 105

Senior Manager
Joined: 18 Jan 2010
Posts: 257

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An entrepreneurship competition requires registering teams to have 3 [#permalink]

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28 May 2016, 08:53
1
KUDOS
Bunuel wrote:
An 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

At least one Tech member is reqd.
Let us find out the tital ways in which 3 people can be selected (without any restrictions). Then we will find out combination when none of techies is selected.

At least once = (Total - None selected)

Total ways in which a team of 3 can be selected out of 13 persons: 13c3 = $$\frac{13!}{3!10!}$$ = $$\frac{11*12*13}{6}$$ = 286

Now suppose we dont want to selcet any Tech guy. Now we have to select 3 persons out of 9 persons: 9c3 = $$\frac{9!}{3!6!}$$
= $$\frac{9*8*7}{6}$$ = 84

Answer: 286 - 84 = 202. E is the answer

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Re: An entrepreneurship competition requires registering teams to have 3 [#permalink]

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25 Sep 2017, 12:25
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Re: An entrepreneurship competition requires registering teams to have 3   [#permalink] 25 Sep 2017, 12:25
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# An entrepreneurship competition requires registering teams to have 3

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