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

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

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Updated on: 27 Feb 2019, 04:20
1
11
00:00

Difficulty:

95% (hard)

Question Stats:

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

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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.
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Re: Question of the Week- 25 (For a meeting, 9 speakers must be .........)  [#permalink]

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06 Dec 2018, 00:03
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
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Re: Question of the Week- 25 (For a meeting, 9 speakers must be .........)  [#permalink]

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02 Dec 2018, 10:44
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
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Re: Question of the Week- 25 (For a meeting, 9 speakers must be .........)  [#permalink]

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30 Nov 2018, 06:52
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 = $$\frac{7}{11*10}$$
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Re: Question of the Week- 25 (For a meeting, 9 speakers must be .........)  [#permalink]

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02 Dec 2018, 00:25
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.
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Re: Question of the Week- 25 (For a meeting, 9 speakers must be .........)  [#permalink]

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02 Dec 2018, 07:08
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
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Re: Question of the Week- 25 (For a meeting, 9 speakers must be .........)  [#permalink]

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05 Dec 2018, 00:01

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.

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Re: Question of the Week- 25 (For a meeting, 9 speakers must be .........)  [#permalink]

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26 Dec 2018, 17:48
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
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