
#1
Oct2905, 04:23 PM

P: 10

I have to prove that for 0<=k<=n that nCk(n choose k) is a natural number. I try by induction but i end up with nm/m+1 must be a natural number which i don't know how to prove, is there another way besides induction? Could someone help me out? Thanks, Jack




#2
Oct2905, 04:31 PM

P: 696

nCk is the number of subsets of cardinality k of, say, {1, 2, ..., n}, which is clearly a natural number (and if this is not your definition of nCk, prove that "your" definition is equivalent to "my" formulation).




#3
Oct2905, 04:50 PM

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P: 3,149

A third approach is to observe that in
[tex]\binom {n}{r} = \frac {n \times (n1) \times \cdot \cdot \cdot \times (nr+1)}{r \times (r1) \cdot \cdot \cdot \times 1}[/tex] the numerator is a product of r successive integers. Therefore, at least one factor is divisible by r, r 1, r  2 etc. so the ratio is a natural number. 



#4
Oct3005, 05:07 AM

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P: 9,398

nCk is a natural number
You know that there is an easy way to write nCk as the sum of two "smaller" combinations, right? Cos then it is inductively a trivial proof.




#5
Oct3005, 06:15 AM

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P: 3,149





#6
Oct3005, 11:20 AM

P: 10

Thanks for all the suggestions guys, could i just ask what is the simpler way of expressing nCr matt? Thanks




#7
Oct3005, 03:24 PM

P: 367

[tex]\binom {n}{r} = \binom {n1}{r1}+\binom{n1}{r}[/tex]
So long as [tex]1 \leq r \leq n1[/tex]. For the cases where this doesn't hold, it's trivial to prove nCr is in N. 



#8
Oct3105, 07:45 AM

P: 534

[tex]\binom{n}{r} = \frac{n!}{r!(nr)!}[/tex] 



#9
Oct3105, 08:08 AM

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jackbauer you should compare the [itex]\binom {n}{r} = \binom {n1}{r1}+\binom{n1}{r}[/itex] expression with Pascal's triangle if you're familiar with it. 



#10
Nov1805, 02:31 PM

P: 28

How would u prove it using induction? any ideas? hints? thx u




#11
Nov1805, 02:42 PM

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P: 1,996





#12
Nov1805, 03:47 PM

P: 28





#13
Nov1805, 04:32 PM

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P: 1,996

Do induction on n. Read the thread in the probability section for more ideas.




#14
Nov1805, 05:35 PM

P: 28

i am trying to prove that it would be a natural number. am i right in this prove?
nCr is n element of N for every o<= r<= n. Suppose that a given r, all the nCr are nutural numbers then since {n+1}Cr = nCr + nC{r1} it follows that the {n+1}Cr are natural numbers for all n. Hence, by inducion, nCr is a natural number for all n and all r. can some one tell me if this prove is correct or can some one help me make it better? thank you so much 



#15
Nov1805, 06:45 PM

P: 943

i recently saw an extremely slick proof of this, but i can't remember where. i think it was a book in the school library. anyway now i'm tortured. the proof said multiply the numerator & denominator by the same thing, then you get [some expression]. end of proof.




#16
Nov1805, 07:53 PM

P: 28

the problem is that i have to proove this by induction....can some one plz review wat i post before and tell me if that prove is right ...thx so much




#17
Nov1905, 12:04 AM

PF Gold
P: 1,059

This problem has come up in this forum over and over again, but I can not now find any references. It is not, as a general rule, easy to prove it by induction. Other methods are better. But using the form as shown by soulflyfgm it can be done. This is similar to using pascal's triangle as a way of obtaining the terms.




#18
Nov1905, 08:50 AM

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P: 1,996

Do you have an easy way to fix this? You could consider primes one at a time. You can count the number of times a prime p appears in a sequence of r numbers. if [] is the greatest integer function, 1*2*3*...*r will be divisible by p to the power [r/p]+[r/p^2]+[r/p^3]+... and no higher power. Any sequence of r integers, no matter where you start, will be divisible by p to this power (possible higher). More generally any sequence of m integers will have at least [m/k] multiples of k in it. So you end up with an integer. soulflyfgm:{n+1}Cr = nCr + nC{r1} doesn't hold for all values of r. 


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