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There are five empty chairs in a row. If six men and four women are wa

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There are five empty chairs in a row. If six men and four women are wa  [#permalink]

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New post Updated on: 18 Oct 2018, 13:23
5
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A
B
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E

Difficulty:

  25% (medium)

Question Stats:

82% (01:50) correct 18% (02:01) wrong based on 114 sessions

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There are five empty chairs in a row. If six men and four women are waiting to be seated, what is the probability that the seats will be occupied by two men and three women?

A) 4/21
B) 5/21
C) 8/21
D) 9/21
E) 11/21


Source: Platinum GMAT
Difficulty Level: Medium

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Originally posted by SajjadAhmad on 14 Jan 2017, 07:13.
Last edited by SajjadAhmad on 18 Oct 2018, 13:23, edited 1 time in total.
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Re: There are five empty chairs in a row. If six men and four women are wa  [#permalink]

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New post 14 Jan 2017, 07:39
SajjadAhmad wrote:
Source: platinum gmat

There are five empty chairs in a row. If six men and four women are waiting to be seated, what is the probability that the seats will be occupied by two men and three women?

A) 4/21
B) 5/21
C) 8/21
D) 9/21
E) 11/21


\(\frac{(6C2 * 4C3)}{10C5} = \frac{5}{21}\)

Answer will be (B)

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Re: There are five empty chairs in a row. If six men and four women are wa  [#permalink]

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New post 23 Oct 2017, 17:19
Bunuel

can you please explain why in this case we do not multiply the 6C2 by 2? and 4C3 by 3?

is it not usually the case that if two men are seated together they can be seated either as 1M, 2M, or 2M or 1M?
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Re: There are five empty chairs in a row. If six men and four women are wa  [#permalink]

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New post 23 Oct 2017, 23:40
ILoveNargiza wrote:
Bunuel

can you please explain why in this case we do not multiply the 6C2 by 2? and 4C3 by 3?

is it not usually the case that if two men are seated together they can be seated either as 1M, 2M, or 2M or 1M?


First of all, I think you meant 2! and 3! not 2 and 3.

Next, we are not interested in their arrangements on chairs, we just want the seats to be occupied by two men and three women.
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Re: There are five empty chairs in a row. If six men and four women are wa  [#permalink]

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New post 28 Aug 2018, 17:43
Bunuel wrote:
ILoveNargiza wrote:
Bunuel

can you please explain why in this case we do not multiply the 6C2 by 2? and 4C3 by 3?

is it not usually the case that if two men are seated together they can be seated either as 1M, 2M, or 2M or 1M?


First of all, I think you meant 2! and 3! not 2 and 3.

Next, we are not interested in their arrangements on chairs, we just want the seats to be occupied by two men and three women.


@ILoveNargiza
I actually, mistakenly, calculated the permutations (no. of arrangements) and I got it right. What I did was: [(6C2 * 4C3 * 5!)/ 10*9*8*7*6]

1. So, firstly, I think you meant - multiple by 5! and not 2! and 3! individually because we would permute among all people, not just among men and women, if we are to find the total possible arrangements.
2. Secondly, here is the thing, as long as you take the ratio of permutations to permutations or combinations to combinations, you should be fine. Of course, It is not recommended but if you happen to confuse one for the other, as long as you are calculating the same thing (arrangements or selections/combinations) in the numerator and denominator, the answer will be the same.

Hope that addresses your question.
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Re: There are five empty chairs in a row. If six men and four women are wa  [#permalink]

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New post 28 Aug 2018, 19:33
Alternative way to solve this question where we do consider the order.

Let's talk about the first possible order

MMWWW

Then the probabilities of picking that specific order is as follows:

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Re: There are five empty chairs in a row. If six men and four women are wa &nbs [#permalink] 28 Aug 2018, 19:33
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