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Choosing Socks


Date: 05/20/99 at 10:08:44
From: Gabriel Alume
Subject: Evaluating Numbers

What are the steps to evaluate the following:

a. C(n,1)

b. C(n,2)

I know the formula is for C(n,r) ... n!/r!(n-r)!

The answer to question b (from the back of the book) is n(n-1)/2. How 
was this answer discovered if the initial step would be n!/2!(n-2)! ?


Date: 05/20/99 at 11:50:11
From: Doctor Jen
Subject: Re: Evaluating Numbers

Hi,

Well, you know that n! = 1*2*3*4*...*(n-2)*(n-1)*n, where the three 
dots ... represent the numbers between 4 and n-2.

If you write out n!/(n-2)!, you get

1*2*3*...*(n-2)*(n-1)*n
------------------------
1*2*3*...*(n-2)

You see that everything in the denominator cancels, and you are left 
with (n-1)*n in the numerator.

So, n!/2!(n-2)! = (n-1)*n/2, since 2! = 2.

You might like to think about how you could do C(n,2) without using 
the formula. Suppose you have n different-coloured socks, and you want 
to choose two to wear. There are n possibilities for the first sock 
you choose. When you have chosen one and taken it out of the drawer, 
there are n-1 socks remaining, so you have a further n-1 possibilities 
for your second sock. That gives you (n-1)*n possible ways of choosing 
two socks. 

But choosing a red sock and a green sock is the same as choosing a 
green sock and a red sock, so each of your possible pairs of socks has 
actually been counted twice. That means there are half of n*(n-1) 
possible pairs of socks. Do you see? Sometimes working C(n,r) out this 
way is easier than using the formula.

- Doctor Jen, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Permutations and Combinations

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