GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 12 Dec 2019, 06:39 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Question of the Week- 25 (For a meeting, 9 speakers must be .........)

Author Message
TAGS:

### Hide Tags

e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3158
Question of the Week- 25 (For a meeting, 9 speakers must be .........)  [#permalink]

### Show Tags

1
11 00:00

Difficulty:   95% (hard)

Question Stats: 28% (03:08) correct 72% (02:40) wrong based on 81 sessions

### HideShow timer Statistics

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. $$\frac{7}{11*5}$$

_________________

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.
Senior Manager  P
Joined: 13 Jan 2018
Posts: 341
Location: India
Concentration: Operations, General Management
GMAT 1: 580 Q47 V23 GMAT 2: 640 Q49 V27 GPA: 4
WE: Consulting (Consulting)
Re: Question of the Week- 25 (For a meeting, 9 speakers must be .........)  [#permalink]

### Show Tags

3
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
Intern  B
Joined: 10 Sep 2018
Posts: 43
Re: Question of the Week- 25 (For a meeting, 9 speakers must be .........)  [#permalink]

### Show Tags

1
+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
Manager  G
Joined: 14 Jun 2018
Posts: 211
Re: Question of the Week- 25 (For a meeting, 9 speakers must be .........)  [#permalink]

### Show Tags

2
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 = $$\frac{7}{11*10}$$
Senior Manager  P
Joined: 03 Mar 2017
Posts: 373
Re: Question of the Week- 25 (For a meeting, 9 speakers must be .........)  [#permalink]

### Show Tags

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.
_________________
--------------------------------------------------------------------------------------------------------------------------
All the Gods, All the Heavens, and All the Hells lie within you.
Manager  G
Joined: 14 Jun 2018
Posts: 211
Re: Question of the Week- 25 (For a meeting, 9 speakers must be .........)  [#permalink]

### Show Tags

warrior1991 wrote:
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.

yea you're right. initially read it as B before A and A before C
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3158
Re: Question of the Week- 25 (For a meeting, 9 speakers must be .........)  [#permalink]

### Show Tags

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.

_________________
Manager  B
Joined: 29 Sep 2016
Posts: 113
Re: Question of the Week- 25 (For a meeting, 9 speakers must be .........)  [#permalink]

### Show Tags

EgmatQuantExpert wrote:

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.

Hi Payal
We can't consider the possibility of B b/n A & C. Re: Question of the Week- 25 (For a meeting, 9 speakers must be .........)   [#permalink] 26 Dec 2018, 17:48
Display posts from previous: Sort by

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