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Updated on: 27 Feb 2019, 04:20
Originally posted by EgmatQuantExpert on 30 Nov 2018, 05:58.
Last edited by EgmatQuantExpert on 27 Feb 2019, 04:20, edited 1 time in total.
Last edited by EgmatQuantExpert on 27 Feb 2019, 04:20, edited 1 time in total.
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E
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Difficulty:
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Question Stats:
34% (03:00) correct
66%
(02:47)
wrong
based on 210
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e-GMAT Question of the Week #25
For a meeting, 9 speakers must be selected to give a speech, one after the other, from a group of 12 people. What is the probability that A, B and C are three among the total 9 speakers selected and B speaks before A and C?
For a meeting, 9 speakers must be selected to give a speech, one after the other, from a group of 12 people. What is the probability that A, B and C are three among the total 9 speakers selected and B speaks before A and C?
- A. \(\frac{7}{12*11*10}\)
B. \(\frac{7}{12*11*5}\)
C. \(\frac{7}{11*3*5}\)
D. \(\frac{7}{11*10}\)
E. \(\frac{7}{11*5}\)
13 Jul 2024, 08:12
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Let's break this into 2 parts: probability that A,B,C will be part of 9 people and probability that B speaks before A and C
Part 1: probability that A,B,C will be part of 9 people
Since A,B,C needs to be in 9 people the 3 spots are taken and rest 6 are up to grab in 12-3c6 i.e 9c6
Total number of ways in which 9 can be selected 12c9.
P(A) = 9c6/12c9 = 9*8*7/12*11*10
Part 2: We need to make sure B speaks before A and C.
Total number of arrangement possible with A,B,C are 3! = 6.
In that B should be front. B A/C C/A = 2!
P(B) = 2/6 = 1/3
now finally P(A)*P(B) = 9*8*7/12*11*10*3 = 7/11*5
Hence option E.
Part 1: probability that A,B,C will be part of 9 people
Since A,B,C needs to be in 9 people the 3 spots are taken and rest 6 are up to grab in 12-3c6 i.e 9c6
Total number of ways in which 9 can be selected 12c9.
P(A) = 9c6/12c9 = 9*8*7/12*11*10
Part 2: We need to make sure B speaks before A and C.
Total number of arrangement possible with A,B,C are 3! = 6.
In that B should be front. B A/C C/A = 2!
P(B) = 2/6 = 1/3
now finally P(A)*P(B) = 9*8*7/12*11*10*3 = 7/11*5
Hence option E.
eswarchethu135
GMAT 2: 640 Q49 V27
Posts: 276
06 Dec 2018, 00:03
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For a meeting, 9 speakers must be selected to give a speech, one after the other, from a group of 12 people. What is the probability that A, B and C are three among the total 9 speakers selected and B speaks before A and C?
Firstly, 9 speakers are selected from 12. This can be done in 12*11*10 ways or \(^{12}P_9\) ways.
Now A,B and C are in the group of 9 people. and B should speak before A and C. This can be done in 7 ways.
A and C be seated after B in the following ways:
B_ _ _ _ _ _ _ _ = \(^8C_2*2!\) (Selecting 2 places among 8 and arranging A and C within their places)
_B_ _ _ _ _ _ _ = \(^7C_2*2!\)
_ _B_ _ _ _ _ _ = \(^6C_2*2!\)
_ _ _B_ _ _ _ _ = \(^5C_2*2!\)
_ _ _ _B_ _ _ _ = \(^4C_2*2!\)
_ _ _ _ _B_ _ _ = \(^3C_2*2!\)
_ _ _ _ _ _B_ _ = \(^2C_2*2!\)
Now B can't be in 8th place as it leaves only one place after B in which both A and C can't sit.
\(\frac{^8C_2*2! + ^7C_2*2! + ^6C_2*2! + ^5C_2*2! + ^4C_2*2! + ^3C_2*2! + ^2C_2*2!}{12*11*10}\)
\(\frac{(7*8)+(6*7)+(5*6)+(4*5)+(3*4)+(2*3)+(1*2)}{12*11*10}\)
\(\frac{56+42+30+20+12+6+2}{12*11*10}\)
\(\frac{168}{12*11*10}\)
\(\frac{7}{11*5}\)
Though it may seem lengthy, if you are confident about the procedure then it should not take more than 1 min 30 seconds to solve this.
OPTION: E
Firstly, 9 speakers are selected from 12. This can be done in 12*11*10 ways or \(^{12}P_9\) ways.
Now A,B and C are in the group of 9 people. and B should speak before A and C. This can be done in 7 ways.
A and C be seated after B in the following ways:
B_ _ _ _ _ _ _ _ = \(^8C_2*2!\) (Selecting 2 places among 8 and arranging A and C within their places)
_B_ _ _ _ _ _ _ = \(^7C_2*2!\)
_ _B_ _ _ _ _ _ = \(^6C_2*2!\)
_ _ _B_ _ _ _ _ = \(^5C_2*2!\)
_ _ _ _B_ _ _ _ = \(^4C_2*2!\)
_ _ _ _ _B_ _ _ = \(^3C_2*2!\)
_ _ _ _ _ _B_ _ = \(^2C_2*2!\)
Now B can't be in 8th place as it leaves only one place after B in which both A and C can't sit.
\(\frac{^8C_2*2! + ^7C_2*2! + ^6C_2*2! + ^5C_2*2! + ^4C_2*2! + ^3C_2*2! + ^2C_2*2!}{12*11*10}\)
\(\frac{(7*8)+(6*7)+(5*6)+(4*5)+(3*4)+(2*3)+(1*2)}{12*11*10}\)
\(\frac{56+42+30+20+12+6+2}{12*11*10}\)
\(\frac{168}{12*11*10}\)
\(\frac{7}{11*5}\)
Though it may seem lengthy, if you are confident about the procedure then it should not take more than 1 min 30 seconds to solve this.
OPTION: E
