Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Combinations for Casting a Play


Date: 11/15/95 at 21:47:58
From: Anonymous
Subject: Permutations & Combinations

I gave this problem to a class of mine, but I now realize it is tad 
more difficult than I first thought.  They however, are dying for an
answer.  Can you help?  I would like to show them the solution using
the fundamental counting principle.

The following shows the lists of people trying out for parts in 
the school play. If someone does not get a part they may try out for 
another if they have signed up. How many possible  casts are there?

Part  Ron     Alice    the butler   Grandpa Jones  Aunt Minnie
                           Jeeves
----------------------------------------------------------------   
      Billy   Susan      Sam        Ralph          Debbie
      Sam     Sally      Ralph      Peter          Alison
      Ralph   Alison     Peter      Joe            Tamara
      Joe     Veronica   Bob                       Ruth
              Debbie                               Lisa


Date: 11/19/95 at 14:44:28
From: Doctor Ken
Subject: Re: Permutations & Combinations

Hello!

Yes, you're right, this problem is kind of hairy.  Let me examine the 
data:

There are three parts for the boys and two parts for the girls.  What we 
can do is figure out how many ways there are to cast the boys' roles, and 
how many ways there are to cast the girls, and then multiply these two 
numbers together to find the total number of casts for the play.  So let's 
look at the girls' roles first, since I bet they're easiest.

Alice     Minnie
-----------------
Susan     Lisa
Sally     Tamara
Veronica  Ruth
Alison    Alison
Debbie    Debbie

What we'll do is first treat the problem as if there were five people 
trying out for Alice and five _different_ people trying out for Minnie; 
then we'll get rid of all the combinations that had one person in two 
different parts.

So let's do it: the number of pairs of roles without elimination is 
5*5=25, and then there are two we'll have to eliminate: (Alison, Alison) 
and (Debbie, Debbie).  So the total number of different combinations for 
the girls' roles will be 25 - 2 = 23.  

The same thinking will apply to the boys' roles, but it will be a little 
more complicated.  Give it a try though, and see what happens.  If 
you're still having trouble, write us back and let us know.  Good luck!

-Doctor Ken,  The Geometry Forum

    
Associated Topics:
High School Permutations and Combinations

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/