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Difficulty with permutation and combination 
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#1
Nov2512, 09:15 AM

P: 106

Right I am having an issue with the proof to permutation, I really can see the [itex]nr1[/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. 


#2
Nov2512, 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*n1*n2*nr+1= 120[/itex]. So I am under the assumption that the [itex] nr+1 [/itex] is the last term in the sequence so to speak, correct or not?



#3
Nov2512, 09:01 PM

P: 439

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 n1 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(r1) possible choices. *So for example, when we had two choices, we had n1 = n(21) So since each choice is indepedent we use the product rule and we obtain n(n1)...(nr+1) = n!/(nr)! Hope that clears up the last term a bit. 


#4
Nov2612, 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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