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/
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