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/
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