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If a class of 10 students has five men, how many ways can the men and
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12 Jan 2010, 07:01
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If a class of 10 students has five men, how many ways can the men and women be arranged in a circle so that no two men sit next to each other? A) 5!4! B) 5!5! C) 4!4! D) 10! E) 10!/5!
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Re: If a class of 10 students has five men, how many ways can the men and
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12 Jan 2010, 07:47
hi ans is A 5!4!.... for notwo men to sit together, either all are in even or odd posn.. fix one at any one posn... then rest four can be fixed in 4! ways... also rest five posn of women can be fixed in 5!.. total ways 5!4!
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Re: If a class of 10 students has five men, how many ways can the men and
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12 Jan 2010, 07:51
Imagine the circle is now a straight line:
_ _ _ _ _ _ _ _ _ _ 1 2 3 4 5 6 7 8 9 10
Let's say that we make a man sits in spot 1. Then, a man must also sit in spots 3, 5, 7, 9. There are 5! ways of arranging the man this way. However, since this is a table (ie: circle), there are 5 ways where the initial man can sit, so we must divide 5! by 5, which equals to 4!.
It follows that a woman must sit in spots 2, 4, 6, 8, 10. There are 5! ways to arrange this.
Therefore, we get 4!5!. So the answer is A) 5!4!



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Re: If a class of 10 students has five men, how many ways can the men and
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12 Jan 2010, 08:00
Since the first position can be fixed by a man or a woman, and then we can have the alternate arrangement of manwomanman........
Shouldn't it be (5!4!)*2? 2  for different choices of the first position.please tell me where i am going wrong.



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Re: If a class of 10 students has five men, how many ways can the men and
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12 Jan 2010, 13:57
You don't have to x2 because you're in a circle.



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Re: If a class of 10 students has five men, how many ways can the men and
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14 Jan 2010, 06:43
thanks for the info
+1 for you



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If a class of 10 students has five men, how many ways can the men and
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02 Oct 2015, 13:10
The pattern has to be MWMWM... or WMWMW...
There are 5! ways of choosing the 5 women, and 5! ways of choosing the 5 men. So we multiply the two, since for each combination of women we have 5! combinations of men. So there are 5!*5! ways of alternating 5 women and 5 men, starting with women.
But what if we started with men? Then we would have the same number of combinations as we do when a woman is first. We multiply by 2. So we have 5!*5!*2 ways of alternating 5 women and 5 men.
Lastly, this is a circular configuration of 10 people, so we divide by 10. Why do we do this? Well, in a circular combination of 3 objects A,B, and C, the configurations ABC, BCA, and CAB are identical. This is because circular configurations look at ORDER, not at ABSOLUTE POSITION. ABC, BCA, and CAB all have the same order (C comes after B, B comes after A...). So in a circular configuration of 3 objects, every configuration can be shifted over 3 times and the order would be the same. The same happens here, so we divide by 10.
The result? \(\frac{5!*5!*2}{10}= 5!4!\)



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Re: If a class of 10 students has five men, how many ways can the men and
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23 Mar 2018, 03:34
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Re: If a class of 10 students has five men, how many ways can the men and
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