Combinations for Casting a PlayDate: 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 |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/