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Subject: "Prove n choose r"     Previous Topic | Next Topic
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Proof Pudding
guest
Feb-13-02, 04:22 PM (EST)
 
"Prove n choose r"
 
   Hi all, I have been ask to look at the problem of proving by induction that n C r = (n!)/((n-r)!r!) ,
0<= r <=n


And I dont know where to start?? For the basis do I take n as being 0, if so r must = 0, there for the basis says n C r = 0 C r = 1.

But what shall the inductive step consist of? If I take (n+1) C r will this work, because r isnt changing in this step so will it be a propor proof???

Please anybody help I am a bit'stuck.

Thanks.


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alexb
Charter Member
681 posts
Feb-13-02, 04:26 PM (EST)
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1. "RE: Prove n choose r"
In response to message #0
 
   >But what shall the inductive step consist of?

Assume you proved the formula for all n less than N and all r not greater than n. In other words, assume you know nCr as long as n < N. Now prove the formula for n = N and all r <= N.


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Sumudu
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Mar-25-02, 11:05 PM (EST)
 
2. "RE: Prove n choose r"
In response to message #0
 
   think of pascal's triangle


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