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MathCounts Problem


Date: 07/18/97 at 17:50:45
From: Wen Li
Subject: Competition problem

Hello Dr. Math,

Here is a question that was on the 1995-96 MathCounts state 
competition. I have the answer, but I do not know how to get it. 

Two boys and four girls are officers of the Math Club. When the 
photographer takes a picture for the school yearbook, she asks the 
club's six officers and the faculty sponsor to sit in a row with the 
faculty sponsor in the middle and the two boys not next to one 
another. How many different seating arrangements are possible?

I am sorry but I really have no idea on how to approach this problem. 
Could you please show me step by step the solution?

Sincerely,
Wen Li  


Date: 07/18/97 at 19:38:02
From: Doctor Anthony
Subject: Re: Competition problem

You have 3 seats either side of the faculty sponsor, and we need 
consider only the six seats of the officers.  The number of 
arrangements of these six people with no restrictions on where they 
sit is 6! = 720.

If the two boys are a and b, then if we group them as (ab) and move 
them about as a single entity, we have only 5 objects to arrange, and 
this can be done in 5! ways = 120 ways.  However the boys could be 
paired as (ba) and again moved around as a single entity, giving a 
further 120 arrangements. So the total number of arrangements when the 
boys ARE together is 240. 

If now we subtract 240 from 720 we shall get the number of 
arrangements when they are NOT together.

   Number of arrangements with boys not together =  720 - 240

                                                 =  480

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   


Date: 07/19/97 at 17:40:49
From: Wen Li
Subject: Wrong answer

Hello Dr. Math,

When I went back to check the answer for this problem, the answer was 
528 arrangements, not 480. I asked Dr. Math this question because I 
wanted to learn the step by step way to answer this question. Could 
Dr. Anthony give me a detailed solution that will result in the answer 
being 528 arrangements?

Thank you for your time.

Sincerely,
Wen Li 


Date: 07/20/97 at 05:25:28
From: Doctor Anthony
Subject: Re: Wrong answer

Yes. There are 48 extra arrangements with the boys not sitting 
together. These occur with the boys seated either side of the sponsor. 
With a on his left and b on his right, the 4 girls can be arranged in 
4! = 24 ways.  And now swapping a and b over so that a is on the right 
and b on the left of the sponsor, we get another 24 arrangements of 
the girls. This gives 24 + 24 = 48 extra arrangements with the boys 
not sitting together. Add this number to the 480 already found and you 
get 528.

If you prefer, you can also get the answer by considering the possible 
arrangements by listing.   In the diagram below, the O represent seats 
of the boys and * represents the seats of the girls.  | is the seat of 
the sponsor. By the arguments given above, for any pair of seats 
occupied by the boys, there will be 48 arrangements of the girls.

     O * O | * * *     the right hand boy could also occupy any seat 
                       to the right of where he is, giving 4 
                       positions.  With each of these there are a  
                       total of 48 arrangements of the girls.  
                       Total 4 x 48

     * O * | O * *     the right hand boy could occupy any seat to the  
                       right of his present position, giving 3  
                       positions. With each of these there are 48  
                       arrangements of the girls, giving a total of  
                       3 x 48.

     * * O | O * *     The right hand boy could occupy any seat to the  
                       right of his present position, giving 3 
                       positions, and a further  3 x 48  arrangements.

     * * * |O * O      There is no other position for the right hand 
                       boy, so this gives 1 x 48 arrangements.

The total of all arrangements with the boys not sitting together is:

          (4 + 3 + 3 + 1) x 48  =  11 x 48

                                =  528.

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Permutations and Combinations

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