Question of the Week- 25 (For a meeting, 9 speakers must be .........)

Question

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?

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}

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qwerty0311bc · Accepted Answer

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.

eswarchethu135 · Answer

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

pandeyashwin · Answer

No of ways of selecting 9 people which includes A,B & C = 9c6 =  \frac{(9*8*7)}{(3*2)} 
This combination includes cases such as A-B-C , C-A-B , B-A-C etc. ( "-" indicates order in which they will speak)
ABC can be arranged in 3! ways. One of those ways is B-A-C. Divide the combination obtained by 3! to remove the arrangement.

\frac{(9*8*7)}{(3*2)3!}  = A 
Total no of ways to select 9 people = 12c9 =  \frac{12*11*10}{3*2}  = B

Probability = A/B =  \frac{7}{11*10}

warrior1991 · Answer

IMO B

Total to be selected =9 
A,B, and C have to be selected . So we pick them and now we have to select 6 more out of the remaining 9.

Ways to select 6 out of 9=9C6

Now we have to check the order in which they can speak . It is mentioned that B will speak before A and C. So order can be BCA or BAC.

ABC as 1 group and rest 6 as 2nd group. So order of speaking will be 7! * (BCA or BAC.).

Possible number of ways= 9C6*7!*2!(A and C can be adjusted in 2! ways)

Total Number of ways = 12C9*9!

Solving we get B as answer.

Hope that helps.

pandeyashwin · Answer

yea you're right. initially read it as B before A and A before C

NCRanjan · Answer

+E
IMO

Select 6 from 9 9C6.

Lets talk about ordering now. question is asking about ordering now B has to speak before A and C ( it does not say that B speaks immediately before A and C . There can be cases like B ,X,Y,Z ,A ,C .

There are only three possibilties B can be before A and C , B can be after A and C or B can be in between A and C . so in exactly 1/3 of the cases B will be ahead of A and C

So favourable outcomes where B is ahead of A and C =9c6/3
Total outcomes - is 12c9

so probability is 7 / 5*11 Option E

EgmatQuantExpert · Answer

Solution 
  Given:  
 • 9 speakers must be selected from a group of 12 people

To find:  
 • The probability that A, B and C are three among the total 9 speakers selected and B speaks before A and C

Approach and Working:   
 Total cases:

• The number of ways of selecting 9 people from a group of 12 =  ^{12}C_9 
• And they can be arranged in 9! ways

Thus, total number of cases =  ^{12}C_9 * 9!

Favourable cases: 
 • Since, A, B and C are three among the 9 selected people, the remaining 6 people must be selected from the group of 9 people. 
 o This can be done in  ^9C_6  ways
o And of the total number of arrangements of these 9 people, 
  B speaks after A and C in  \frac{1}{3}^{rd}  of the cases =  \frac{9!}{3} 
 B speaks between A and C in  \frac{1}{3}^{rd}  of the cases =  \frac{9!}{3} , and
 B speaks before A and C in  \frac{1}{3}^{rd}  of the cases =  \frac{9!}{3}

• Thus, total number of favorable cases =  ^9C_6 * \frac{9!}{3}

Therefore, the probability that A, B and C are three among the total 9 speakers selected and B speaks before A and C =  \frac{^9C_6 * 9!}{3 * ^{12}C_9 * 9!} = \frac{9 * 8 * 7}{(12 * 11 * 10 * 3)} = \frac{7}{11*5}

Hence the correct answer is Option E.

Answer: E

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AKY13 · Answer

Hi Payal
We can't consider the possibility of B b/n A & C.

Budhaditya_Saha · Answer

I guess this is the best and easiest approach to the question.

Posted from my mobile device

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Question of the Week- 25 (For a meeting, 9 speakers must be .........)

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Question of the Week- 25 (For a meeting, 9 speakers must be .........)

Prep Toolkit

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