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Permutations question

 
 
Picture of Heidi McInnes
Permutations question
by Heidi McInnes - Wednesday, 20 March 2013, 9:09 PM
 

I have given the following question to my students:

"Connor hosts a morning radio show in Edmonton.  To advertize the local University team, he is holding a contest at a local mall.  He spells out PANDAS with letter tiles.  Then he turns the tiles face down and mixes them up.  He asks Gabe to arrange the tiles in a row and turn them face up.  If the row of tiles spells PANDAS, Gabe will win a new car.  Determine the probability that Gabe will win a new car.  (express the probability as a reduced fraction)"

 

now, my answer is 1/360.   I get the 360 by going 6!/2!  because of the two identical A's.  I have one as the numerator because there is only 1 way to spell PANDAS, however many of my students are making the numerator 2 because they are taking into account the identical A's in the arrangement of PANDAS.  Can anyone explain to me why or why not the numerator should be 1   or a 2??

 

Thanks for your help.  (I can't wait to be done this unit)

 

Picture of Steve Stahl
Re: Permutations question
by Steve Stahl - Wednesday, 20 March 2013, 11:00 PM
 

Ha ha...good ol' permutations eh?!

There really is two ways to solve this question.  

-Let's say you ignore the double 'A'.  Then you could say that there are 2 favorable ways to arrange the word PANDA out of a total of 6! ways to arrange all the letters. This gives you 2/6! which is 1/360 as well.

-I believe the way you taught it is better.  Given that there are repitions it would be best to exclude the repition early.  If you exclude the repition by dividing the total number of arrangements by 2!  (ie. 6!/2!)...then because the repitions are excluded there is only 1 favorable outcome.

this is a tough one too try to explain with text! hope this helps you!!

Steve

 

The way I see it the numerator could be a 2 if you ignore the fact that there is a double 'A' in pandas...  hence 2/(6!).  if you ignore the double 'A' then you have 2 favorable outcomes out of 6! total outcomes.

A better solution however would be to consider the double 'A' into the calculation and 'divide' it out....that's