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An entrepreneurship competition requires registering teams to have 3
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Bunuel
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An entrepreneurship competition requires registering teams to have 3 [#permalink] New post  26 May 2016, 11:33
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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
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nahid78
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An entrepreneurship competition requires registering teams to have 3 [#permalink] New post  Updated on: 06 Mar 2018, 09:26
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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

=286 - 84

= 202 (E)
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Originally posted by nahid78 on 26 May 2016, 13:27.
Last edited by nahid78 on 06 Mar 2018, 09:26, edited 1 time in total.
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sudhirmadaan
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An entrepreneurship competition requires registering teams to have 3 [#permalink] New post  28 May 2016, 05:55
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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fantaisie
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An entrepreneurship competition requires registering teams to have 3 [#permalink] New post  28 May 2016, 06:24
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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

Answer: E

Are there any shortcuts to avoid doing all the scenarios separately?
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adiagr
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An entrepreneurship competition requires registering teams to have 3 [#permalink] New post  28 May 2016, 07:53
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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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louisbharnabas
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Re: An entrepreneurship competition requires registering teams to have 3 [#permalink] New post  05 Mar 2018, 07:29
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


I initially did 4C1*12C2 (=264), which is not in the options; therefore, used 13C3-9C3 (=202), which is the correct answer.

Not sure what I am doing wrong with the first method. Can someone tell me what i am doing wrong with 4C1*12C2??

Regards,
Louis
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sobby
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Re: An entrepreneurship competition requires registering teams to have 3 [#permalink] New post  05 Mar 2018, 08:00
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louisbharnabas 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


I initially did 4C1*12C2 (=264), which is not in the options; therefore, used 13C3-9C3 (=202), which is the correct answer.

Not sure what I am doing wrong with the first method. Can someone tell me what i am doing wrong with 4C1*12C2??

Regards,
Louis


take a simple example:A,B,C and D --4 tech guys
4C1*12C2
A (one mem out of 4) -- BD(two mem out of rest 12 ) --can be a team
B (one mem out of 4) -- AD(two mem out of rest 12) --can also be a team , but both team are same .
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louisbharnabas
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Re: An entrepreneurship competition requires registering teams to have 3 [#permalink] New post  05 Mar 2018, 08:07
sobby wrote:
louisbharnabas 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


I initially did 4C1*12C2 (=264), which is not in the options; therefore, used 13C3-9C3 (=202), which is the correct answer.

Not sure what I am doing wrong with the first method. Can someone tell me what i am doing wrong with 4C1*12C2??

Regards,
Louis


take a simple example:A,B,C and D --4 tech guys
4C1*12C2
A (one mem out of 4) -- BD(two mem out of rest 12 ) --can be a team
B (one mem out of 4) -- AD(two mem out of rest 12) --can also be a team , but both team are same .


Dear Sobby,

Oh yes. You are right. Now i understand.

Thank you.

Regards,
Louis
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