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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
```

```
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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