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Seating People in a Row


Date: 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/

    
Associated Topics:
High School Permutations and Combinations

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