April 29th, 2011, 03:16 PM  #1 
Newbie Joined: Apr 2011 Posts: 26 Thanks: 0  Permutation
Two families, each have a husband, a wife, two sons and a daughter, are planning to go to a baseball game and sit together in the same row. In how many ways can they be seated if all the males are to sit together and all the females to sit together? Males sitting together = 6! = 6*5*4*3*2*1 = 720 Females siting together = 4! = 4*3*2*1 = 24 Am i correct by saying this ?? 
April 29th, 2011, 03:26 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Permutation
Yes, also consider they may be arranged: MMMMMMFFFF or FFFFMMMMMM 
April 29th, 2011, 03:32 PM  #3 
Newbie Joined: Apr 2011 Posts: 26 Thanks: 0  Re: Permutation
So how would i include them all together ... by adding the outcomes together ??? 720+24 = 744 or 10*9*8*7*6*5*4*3*2*1 = 3628800 
April 29th, 2011, 03:38 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Permutation
No, you multiply, then double for the two arrangements I gave above: N = 2(6!4!) = 34560 

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