In how many ways can we fill seven numbered seats with three : DS Archive
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# In how many ways can we fill seven numbered seats with three

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Manager
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In how many ways can we fill seven numbered seats with three [#permalink]

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08 Aug 2003, 00:17
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In how many ways can we fill seven numbered seats with three men and four women by selection from six men and five women?
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08 Aug 2003, 01:48
First select 3 men and 4 women = 6C3*5C4=100 selections
Each seven-person selection should be placed in 7 places = 7!= 5040

total 504000
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08 Aug 2003, 08:40
stolyar wrote:
First select 3 men and 4 women = 6C3*5C4=100 selections
Each seven-person selection should be placed in 7 places = 7!= 5040

total 504000

Stolyar, since the women are all same and men are all same, the multiplication factor should be

7!/3!*4! = 35 and therefore the total ways should be 100*35 = 3500

What do you think?
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08 Aug 2003, 10:51
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Respect,

KL

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08 Aug 2003, 13:59
prashant wrote:
stolyar wrote:
First select 3 men and 4 women = 6C3*5C4=100 selections
Each seven-person selection should be placed in 7 places = 7!= 5040

total 504000

Stolyar, since the women are all same and men are all same, the multiplication factor should be

7!/3!*4! = 35 and therefore the total ways should be 100*35 = 3500

What do you think?

I couldn't understand when you said "since the women are all same and men are all same, the multiplication factor should be

7!/3!*4! = 35 and therefore the total ways should be 100*35 = 3500"

BTW: I got the same answer as Saylor. But, he agreed you are right.
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08 Aug 2003, 22:10
I am not sure of the explanation too. In any such questioon about men/women, we NEVER assume people are same. The explanation for 7!/(4!*3!) would be ok if there were red and white balls. However, for men/women, original solution of Stolyar seems correct. Comments ?
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09 Aug 2003, 04:35
anupag wrote:
I am not sure of the explanation too. In any such questioon about men/women, we NEVER assume people are same. The explanation for 7!/(4!*3!) would be ok if there were red and white balls. However, for men/women, original solution of Stolyar seems correct. Comments ?

The question says that there are 6 men and 5 women, and we have to choose 3 men and 4 women.

The only two "categories" of choice are Men and Women - woman 1 is sitting in chair 1 is not different from woman 2 sitting in chair 1.

Therefore, in order to arrange these choices into 7 chairs, the only way the arrangements will differ is if a certain chair is occupied by a man or a woman.

Therefore, the number of arrangements for 3 men and 4 women are

(3+4)!/3!*4! = 7!/3!*4! = 35

Multiply this by the number of ways in which the choices of men and women were made and you get the answer = 100*35=3500

Hope this helps - I'm not a tutor like Akamai, but I try my best
09 Aug 2003, 04:35
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