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Re: Four women and three men must be seated in a row for a group [#permalink]
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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]
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]
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]
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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]
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]
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Re: Four women and three men must be seated in a row for a group [#permalink]
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