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.
Please write back if you have any more questions about this.

- Doctor Douglas, The Math Forum
  http://mathforum.org/dr.math/