Choosing a Duet, Lead Singer/BackupDate: 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! |
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