Seating People in a RowDate: 12/16/95 at 18:58:1 From: Anonymous Subject: Permutation and Combination Find the numbers of ways in which 4 boys and 4 girls can be seated in a row of 8 seats if they sit alternately and if there is a boy named John and a girl named Susan amongst this group who cannot be put in adjacent seats. Solutions: Case 1 : 3*3! * 2 2*(4*4*3*3*2*2*1*1) = 1152 Case 2 : 3! * 2 * 2 Case 3 : 2 * 2 * 3! Case 4 : 2 * 2 * 3! Total ways : 1152- 2 * ( Case1 + case2+case3+case4) However the answer in my textbook is 648 ways and I didn't get the result. Date: 5/29/96 at 15:16:58 From: Doctor Anthony Subject: Re: Permutation and Combination It is best to find the number of ways with John and Susan sitting together and subtract this from the total without the restriction. Let us bracket J and S and consider it to be a single entity. So a possible arrangement is (B=boy, G=girl) (JS) B G B G B G number of arrangements is 7 x 3! x 3! = 252 (SJ) G B G B G B number of arrangements is 7 x 3! x 3! = 252 So total with JS together is 504 Now unrestricted arrangements are B G B G B G B G = 4! x 4! = 576 or G B G B G B G B = 4! x 4! = 576 and this total is 1152. To satisfy condition that J and S are NOT together we must subtract 504 Required number of ways = 1152 - 504 = 648 -Doctor Anthony, The Math Forum Date: 5/31/96 at 11:3:7 From: Doctor Anthony Subject: Re: Permutation and Combination Let me add a little detail to the answer I gave for this question. I start by bracketing John and Susan together, and denote this by (JS). I treat this bracket as a single entity, and it could occupy one of seven positions in the line-up. (There are six other people to be arranged in the line). Now the three remaining boys could be arranged in 3! (=6) ways, and the three girls also in 3! ways. So with the bracket (JS) plus six others, the total number of acceptable arrangements (i.e. with boys and girls alternating) is 7 x 3! x 3! = 252. We could of course have bracketed John and Susan in the order (SJ), and again this could give a total of 7 x 3! x 3! = 252 ways. So the total with John and Susan together either as JS or SJ is 504. The total of possible arrangements with boys and girls alternating and with a boy at the start is 4! x 4! = 576. And the total with a girl at the start is also 4! x 4! = 576 giving a grand total of 1152. Now to satisfy the condition that John and Susan are NOT together we subtract the number of ways when they ARE together (=504) Required number of ways = 1152 - 504 = 648 -Doctor Anthony, The Math Forum Date: 08/01/2003 at 01:04:50 From: Marian May Subject: Statistics, combination and permutation In how many ways can 4 boys and 4 girls be seated at a round table alternately? Date: 08/11/2003 at 10:14:35 From: Doctor Anthony Subject: Re: Statistics, combination and permutation Fix one boy and arrange the other 3 boys in 3! ways. Arrange the 4 girls in 4! ways in the gaps between the boys. Total arrangements = 3! x 4! = 6 x 24 = 144 - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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