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Four women and three men must be seated in a row for a group

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GreenlightTestPrep
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Four women and three men must be seated in a row for a group [#permalink] New post   06 Jun 2017, 13:05
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Four women and three men must be seated in a row for a group photograph. If no two men can sit next to each other, in how many different ways can the seven people be seated?

A) 240
B) 480
C) 720
D) 1440
E) 5640
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Four women and three men must be seated in a row for a group [#permalink] New post   07 Jun 2017, 15:02
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GreenlightTestPrep wrote:
Four women and three men must be seated in a row for a group photograph. If no two men can sit next to each other, in how many different ways can the seven people be seated?

A) 240
B) 480
C) 720
D) 1440
E) 5640


Take the task of arranging the 7 peopl and break it into stages.

Stage 1: Arrange the 4 women in a row
We can arrange n unique objects in n! ways.
So, we can arrange the 4 women in 4! ways (= 24 ways)
So, we can complete stage 1 in 24 ways

IMPORTANT: For each arrangement of 4 women, there are 5 spaces where the 3 men can be placed.
If we let W represent each woman, we can add the spaces as follows: _W_W_W_W_
So, if we place the men in 3 of the available spaces, we can ENSURE that two men are never seated together.

Let's let A, B and C represent the 3 men.

Stage 2: Place man A in an available space.
There are 5 spaces, so we can complete stage 2 in 5 ways.

Stage 3: Place man B in an available space.
There are 4 spaces remaining, so we can complete stage 3 in 4 ways.

Stage 4: Place man C in an available space.
There are 3 spaces remaining, so we can complete stage 4 in 3 ways.

By the Fundamental Counting Principle (FCP), we can complete the 4 stages (and thus seat all 7 people) in (24)(5)(4)(3) ways (= 1440 ways)

Answer:
Show: ::
D


Note: the FCP can be used to solve the MAJORITY of counting questions on the GRE So be sure to learn this technique.

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Runnyboy44
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Re: Four women and three men must be seated in a row for a group [#permalink] New post   22 Aug 2018, 00:49
May I know why is it when we place the men, we don't follow the reasoning where since there are three men, there are three men to choose from, hence 3x, followed by 2x and 1x. Instead, we follow the reasoning of the available chairs instead i.e. 5 chairs, then 4, then 3. I agree with your way as it seems correct but I can't grasp the logic behind it. Thanks for helping me to better understand.
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Re: Four women and three men must be seated in a row for a group [#permalink] New post   22 Aug 2018, 16:12
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Runnyboy44 wrote:
May I know why is it when we place the men, we don't follow the reasoning where since there are three men, there are three men to choose from, hence 3x, followed by 2x and 1x. Instead, we follow the reasoning of the available chairs instead i.e. 5 chairs, then 4, then 3. I agree with your way as it seems correct but I can't grasp the logic behind it. Thanks for helping me to better understand.


Say there are 9 chairs so then there is a blank space to the left of a woman and a blank space to the right. Like below

_ _ _ _ _ _ _ _ _

So we can place women on chair 2 chair 4....chair 8.

Chair 2: 4 woman available for seating

Chair 4: 3 woman available for seating

Chair 6: 2 woman available for seating

Chair 8: 1 woman available for seating

So total ways = \(4 \times 3 \times 2 \times 1= 4!=24\).

Now blanks spaces available for man_1 = 5; man_2= 4; man_3=3

Total ways is \(5 \times 4 \times 3= 60\).

In the first case we are arranging the women in the second case we are arranging the blanks. When we say no two man can sit together the positions of women are fixed: i.e. 2, 4, 6, 8. Position of men are not fixed. So a viable arrangement can be:

M-W-W-M-W-M-W vs W-W-M-W-M-W-M.

If you had only 3 places for men to sit then the only viable combination would have been \(3 \times 2 \times 1= 3!\) just like the women case but here we have 5 places and 3 men, hence \(5 \times 4 \times 3=60\).


Note this is also called arranging m things in n places and it is represented by:

\(P^{n}_{m}=\frac{n!}{(n-m)!}\)

So 5 chairs 3 men: \(P^{5}_{3}=\frac{5!}{(5-3)!}= 5 \times 4 \times 3\)
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Re: Four women and three men must be seated in a row for a group [#permalink] New post   30 Oct 2018, 11:55
Why do we have to find the ways to arrange the women first? Would it be possible to solve this problem by finding all the different ways of arranging the 7 people then subtract out the number of arrangements that violate the restriction?
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Re: Four women and three men must be seated in a row for a group [#permalink] New post   30 Oct 2018, 12:11
msawicka wrote:
Why do we have to find the ways to arrange the women first? Would it be possible to solve this problem by finding all the different ways of arranging the 7 people then subtract out the number of arrangements that violate the restriction?


There's only one way to find out . . . :-D
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Re: Four women and three men must be seated in a row for a group [#permalink] New post   16 Apr 2020, 07:28
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There is another way to consider this question:
In case one: W_W_W_W
In case two: _W_W_WW
In case three: WW_W_W_

This would ensure that no two men are sitting together; it also complicates the above solution, as the solution would be 3*4!*3!, or 432.
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Re: Four women and three men must be seated in a row for a group [#permalink] New post   27 Apr 2020, 07:21
Question wrote:
There is another way to consider this question:
In case one: W_W_W_W
In case two: _W_W_WW
In case three: WW_W_W_

This would ensure that no two men are sitting together; it also complicates the above solution, as the solution would be 3*4!*3!, or 432.


There are actually 10 different cases in total.
1. W_W_W_W
2. _WW_W_W
3. WW_W_W_
4. _W_WW_W
5. W_WW_W_
6. _W_W_WW
7. W_W_WW_
8. _WWW_W_
9. _W_WWW_
10. _WW_WW_

Each of the 10 possible configurations can be achieved in (4!)(3!) ways.

Cheers,
Brent
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Re: Four women and three men must be seated in a row for a group [#permalink] New post   19 Sep 2023, 05:54
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