# Homework Help: What's wrong with this

1. Nov 7, 2007

### hazim

[SOLVED] what's wrong with this

nPr= n!/(n-r)! and here in permutation order is important, and in combination
nCr= n!/( r!(n-r)! ) order is not important, so the propability of ana event will be more for combination from permutation, but permutation is combination times r!, so what is wrong with me and not with my question

Last edited: Nov 7, 2007
2. Nov 7, 2007

### sara_87

i dont know who told u that in permutation order is not important because i think it is important...that's the difference between permutation and combination.

3. Nov 7, 2007

### hazim

sorry i was mistaken but now i corrected my question

4. Nov 7, 2007

### sara_87

the PROBABILITY will be more for combination
eg. nCr=7
and nPr= 60
so lets say the number 2 is in both then prob for nCr is 2/7
and for nPr it's 2/60
2/7 is greater than 2/60
am i making any sense?
i'm not sure if that's 100% right.

5. Nov 7, 2007

### hazim

i didn't understand you, i mean if n and r are the same for combination and permutation, the result will be nCr > nPr, that is what i'm not understanding, but what is r and n in your post?

6. Nov 7, 2007

### sara_87

whatever r and n are...it doesnt matter because u want the probability, so if nPr>nCr, then the probaility of nCr is > prob of nPr.