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### Rugby Order

```Date: 08/23/2003 at 00:45:56
From: Ellie
Subject: Rugby order

If there are 8 players and 5 positions, making 6720 different
arrangements possible, how is that number reduced if 2 of the players
can only play on the very outside right, and 1 can only play in the
middle?

I know that there are 120 ways that they can be arranged without the
3 people who are 'special', but I don't know how to easily work the
rest out.
```

```
Date: 08/23/2003 at 10:04:52
From: Doctor Douglas
Subject: Re: Rugby order

Hi Ellie,

Thanks for writing to the Math Forum.

Let me designate the five players who can play anywhere as group A
(the "anywhere" group). And let's call the two players who are
replacement rightsides by set R, and M the player who can only play
middle.

We simply break down the count (this is the easiest way I know how to
take care of all of the constraints on who can play where):

# = (#,outside from A) + (#,outside from R) + (#,outside from M)
= (#,outside from A) + (#,outside from R) + 0
= (#,out from A, mid from A) + (#,out from A, mid from M)
+ (#,out from R, mid from A) + (#,out from R, mid from M)

Do you see how this lists all of the possibilities?  Notice how
(#,outside from M) is set to zero because that is not a valid team -
the player in M cannot play outside. Similarly, I didn't write terms
of the form (#,...,mid from R). Each of these four remaining terms is
not too difficult:
--others-   out  mid
(#,out from A, mid from A)  = 5 x 4 x 3 x 2A x 1A   note 2A then 1A
(#,out from A, mid from M)  = 5 x 4 x 3 x 2A x 1M
(#,out from R, mid from A)  = 5 x 4 x 3 x 2R x 2A   note 2A's!
(#,out from R, mid from M)  = 5 x 4 x 3 x 2R x 1M

for which the sum total is

# = 5 x 4 x 3 x (2x1 + 2x1 + 2x2 + 2x1) = 5 x 4 x 3 x (10)
= 1200

Things can get very complicated with more intricate constraints.

- Doctor Douglas, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Permutations and Combinations

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