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
_____________________________________________

Choosing a Duet, Lead Singer/Backup


Date: 02/25/2002 at 23:47:53
From: Bryan Smoot
Subject: Statistics probability

This is two-part question is confusing me.

There are 20 singers trying out for a musical. In how many different 
ways can the director choose a duet? 

In how many ways can the director choose a lead singer and a backup?

I thought the answers would be the same for both parts because it 
would always end up being two people chosen, but apparently this is 
not correct. Please help clear up this confusion up for me.

Thanks.


Date: 02/27/2002 at 19:38:56
From: Doctor Tim
Subject: Re: Statistics probability

Hi!

This is a great question because it illuminates a really touchy issue 
in this kind of problem - I'm going to have to remember it for my 
students.

Let's try a simpler case:

Consider if there are only two singers, A and B. How many ways can 
there be a duet? Only one. But there are two ways (AB, BA) to choose a 
lead and backup. That is, in the second situation, ORDER MATTERS.

For a moment, let's do 3 singers.
Duet: AB AC BC. But order doesn't matter. 3 ways.
Lead/backup: AB BA AC CA BC CB. Order does matter. 6 ways.

Will it always be twice as many ways to do lead/backup? Sure. Because 
every duet has two lead/backup configurations, and none of them 
duplicates the lead/backup configurations for any other duet.

SO: how many ways can you choose 2 people out of 20? Let's do lead/
backup FIRST:

20 ways to pick the lead.
For each way, there are 19 ways to pick the backup.
So the answer is 20 * 19 or 380.

Now, from what we did above, we know that the number of duets is half 
that: 190.

NOTE that this way of counting works fine, but it is also the same as 
the combinations formula you probably know (which is for order-
doesn't-matter, the duet situation):

   C(n, k) = n! / [(n-k)! k!]

   C(20,2) = 20! /[(20-2)! 2!] = 20!/(18! 2!) = 20 * 19 / 2 = 190.

It took me YEARS to realize that the whole reason for the (n-k) in the 
denominator was to cancel all but the first few numbers in the n!. 
That is, you don't really want 20!, you just want 20*19. So one way to 
get that is to call it 20!/18! - which cancels all the factors from 18 
down.

I hope this helps!

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


Date: 03/01/2002 at 12:10:12
From: Bryan Smoot
Subject: Statistics probability

Thanks so much for your input and help on this question. Your reply 
provided much clarity to this confusing problem. Keep up the good 
work!
    
Associated Topics:
High School Discrete Mathematics
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/