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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

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?

Sincerely,
Wen Li
```

```
Date: 07/20/97 at 05:25:28
From: Doctor Anthony

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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