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Find the number of ways in which 5 men and 4 women can be

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Find the number of ways in which 5 men and 4 women can be [#permalink] New post 26 May 2006, 05:43
Find the number of ways in which 5 men and 4 women can be seated round a table so that no two women are together.
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 [#permalink] New post 26 May 2006, 06:07
Total number of ways to sit 5 men in 4 women near the table=5C9+4C9=252

Number of ways to sit 2 women near each other=9 (this may be a mistake :) )

Total number of ways to sit so that no 2 women sit near each other=252-9=243
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 [#permalink] New post 26 May 2006, 07:11
IMO
it should be MMWMWMWMW=4!x3!
MWMWMWMWM=4!x3!
WMWMWMWMM=4!x3!
total-144x3=432
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 [#permalink] New post 26 May 2006, 12:02
There is no OE. But the following may help you solve such problems ---

The number of ways of arranging n persons along a round table so that no person has the same two neighbours is(n-1)!/2
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 [#permalink] New post 26 May 2006, 12:03
Source --- http://www.ilovemaths.com/3permcirc.htm
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 [#permalink] New post 27 May 2006, 10:02
mahesh004 wrote:
There is no OE. But the following may help you solve such problems ---

The number of ways of arranging n persons along a round table so that no person has the same two neighbours is(n-1)!/2


That is the formula when we treat anti-clockwise and clockwise arrangements identical.

Should it not be

five men in (5-1)!

4 w in 5 spots 5P4
4! x 5!

:?
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 [#permalink] New post 27 May 2006, 12:25
but your answer is same
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 [#permalink] New post 27 May 2006, 12:49
Yes, the formula for the arrangement in a circle is (x-1)!, so it is 4!x5!
  [#permalink] 27 May 2006, 12:49
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