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# Five people, Ada, Ben, Cathy, Dan, and Eliza, are lining up for a

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Joined: 13 Oct 2015
Posts: 17
Five people, Ada, Ben, Cathy, Dan, and Eliza, are lining up for a  [#permalink]

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Updated on: 11 Jan 2017, 10:24
2
00:00

Difficulty:

45% (medium)

Question Stats:

60% (01:29) correct 40% (02:10) wrong based on 70 sessions

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Five people, Ada, Ben, Cathy, Dan, and Eliza, are lining up for a photo. What is the probability that none of the three girls will stand next to one another?

(A) 0.1
(B) 0.2
(C) 0.3
(D) 0.4
(E) 0.5

Originally posted by DrAB on 07 Sep 2016, 09:32.
Last edited by Bunuel on 11 Jan 2017, 10:24, edited 1 time in total.
Edited the options.
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Re: Five people, Ada, Ben, Cathy, Dan, and Eliza, are lining up for a  [#permalink]

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07 Sep 2016, 10:26
2
DrAB wrote:
Five people, Ada, Ben, Cathy, Dan, and Eliza, are lining up for a photo. What is the probability that none of the three girls will stand next to one another?

(A) 0,1
(B) 0,2
(C) 0,3
(D) 0,4
(E) 0,5

Out of 5 , then the only position the three girls can take are 1, 3 and 5th so they are have 3! ways of standing in those position and remaining 2 boys can have 2! ways. So Probability = $$\frac{3! * 2!}{5!} =\frac{1}{10}$$
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Re: Five people, Ada, Ben, Cathy, Dan, and Eliza, are lining up for a  [#permalink]

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07 Sep 2016, 18:39
Senthil1981 wrote:
DrAB wrote:
Five people, Ada, Ben, Cathy, Dan, and Eliza, are lining up for a photo. What is the probability that none of the three girls will stand next to one another?

(A) 0,1
(B) 0,2
(C) 0,3
(D) 0,4
(E) 0,5

Out of 5 , then the only position the three girls can take are 1, 3 and 5th so they are have 3! ways of standing in those position and remaining 2 boys can have 2! ways. So Probability = $$\frac{3! * 2!}{5!} =\frac{1}{10}$$

This is a poor-quality question.
How can one tell there are 3 girls?
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Re: Five people, Ada, Ben, Cathy, Dan, and Eliza, are lining up for a  [#permalink]

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07 Sep 2016, 20:19
I did it this way...

BTW we can tell the number of girls (although I though this should be more clear) because, based on the answer choices, we know there is at least ONE way they can stand all separate. So there are 3 girls and 2 boys.

Total number of options they could stay 5!=120

The number of ways for the girls to be separate

3*2*2*1*1=12
12/120=1/10
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Re: Five people, Ada, Ben, Cathy, Dan, and Eliza, are lining up for a  [#permalink]

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07 Sep 2016, 20:42
1
Donnie84 wrote:
Senthil1981 wrote:
DrAB wrote:
Five people, Ada, Ben, Cathy, Dan, and Eliza, are lining up for a photo. What is the probability that none of the three girls will stand next to one another?

(A) 0,1
(B) 0,2
(C) 0,3
(D) 0,4
(E) 0,5

Out of 5 , then the only position the three girls can take are 1, 3 and 5th so they are have 3! ways of standing in those position and remaining 2 boys can have 2! ways. So Probability = $$\frac{3! * 2!}{5!} =\frac{1}{10}$$

This is a poor-quality question.
How can one tell there are 3 girls?

The question says "none of the three girls". Could this mean differently ?
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Re: Five people, Ada, Ben, Cathy, Dan, and Eliza, are lining up for a  [#permalink]

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07 Sep 2016, 20:49
The question says "none of the three girls". Could this mean differently ?[/quote]

Ah, I missed that.
+1 to you.

Still, I think an official question will have a clearer language than this one.
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Re: Five people, Ada, Ben, Cathy, Dan, and Eliza, are lining up for a  [#permalink]

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07 Sep 2016, 21:01
Donnie84 wrote:
The question says "none of the three girls". Could this mean differently ?

Ah, I missed that.
+1 to you.

Still, I think an official question will have a clearer language than this one.[/quote]

It's true that this not in GMAT format since they have used "," instead of ".". Looks like European notation.
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Re: Five people, Ada, Ben, Cathy, Dan, and Eliza, are lining up for a  [#permalink]

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04 Jan 2018, 15:58
Hi All,

We're given 3 girls and 2 boys and told to put them in a line. We're asked for the probability that none of the three girls will stand next to one another. This question can be approached in a number of different ways. Here's how you can use permutations and probability to get to the correct answer:

With 5 people, there are 5! = 120 possible arrangements. Lining up the 5 people, there's just one option for the 3 girls to NOT stand next to one another:

GBGBG = (3)(2)(2)(1)(1) = 12 options

Thus, the probability of the 3 girls NOT standing next to one another is 12/120 = 1/10 = 10% = .1

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Re: Five people, Ada, Ben, Cathy, Dan, and Eliza, are lining up for a &nbs [#permalink] 04 Jan 2018, 15:58
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