Difficulty with permutation and combination


by Taylor_1989
Tags: combination, difficulty, permutation
Taylor_1989
Taylor_1989 is offline
#1
Nov25-12, 09:15 AM
P: 106
Right I am having an issue with the proof to permutation, I really can see the [itex]n-r-1[/itex]
I think the confusion stems because it is in the general term, which throws me a bit, if possible could someone maybe write it in numbers and the underneath write in the general term if not too much trouble. The reason I ask for this is I am trying to understand the binomial expansion, and I have never done permutation or comnations be for, I do understand factorials and how permutation work and combination, but can't get my head around the proof for the formula.

I would like to thank anyone in advance for posting a replie to this post, much appreciated.
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Taylor_1989
Taylor_1989 is offline
#2
Nov25-12, 10:54 AM
P: 106
Okay so I have resized I am being quite vague, when I posted this. So I have been having a look at the proof a going over it so please correct me if I am wrong to what I am about to write: if I have say 5 letter - ABCDE and have 5 place to fill this I would get this type of equation: [itex]5*4*3*2*1= 120[/itex] So I have 120 permutation. In general form this would N=5 and the R=5 [itex]n*n-1*n-2*n-r+1= 120[/itex]. So I am under the assumption that the [itex] n-r+1 [/itex] is the last term in the sequence so to speak, correct or not?
MarneMath
MarneMath is offline
#3
Nov25-12, 09:01 PM
P: 422
I think you are getting it, but, I'm going to go over an informal proof to see if you can follow.

Let us select elements of S in any order. There are n elements within S, for our first selection we have n options. Thus there are n - 1 elements left in S, so we have n-1 for the second choice, and thus we now have n - 2 options left, and henece n - 2 choices for the third option. If we notice this pattern, we can say that for the rth choice there are n-(r-1) possible choices.

*So for example, when we had two choices, we had n-1 = n-(2-1)

So since each choice is indepedent we use the product rule and we obtain n(n-1)...(n-r+1) = n!/(n-r)!

Hope that clears up the last term a bit.

Taylor_1989
Taylor_1989 is offline
#4
Nov26-12, 02:21 PM
P: 106

Difficulty with permutation and combination


Yep, I see where you are coming from. I had the right idea i my head, but couldn't get what I wanted to say on here. Thanks for the input def cleared things up.


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