nCk is a natural number

jackbauer

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 n-m/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

Muzza

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).

Tide

A third approach is to observe that in

[tex]\binom {n}{r} = \frac {n \times (n-1) \times \cdot \cdot \cdot \times (n-r+1)}{r \times (r-1) \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.

matt grime

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.

Tide

It looks like we've pinned down all the permutations of proofs for this one! ;)

jackbauer

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

Moo Of Doom

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

ComputerGeek

I thought that
[tex]\binom{n}{r} = \frac{n!}{r!(n-r)!}[/tex]

shmoe

This is the same. Tide has just cancelled the common factors of n! and (n-r)!.

jackbauer- you should compare the [itex]\binom {n}{r} = \binom {n-1}{r-1}+\binom{n-1}{r}[/itex] expression with Pascal's triangle if you're familiar with it.

soulflyfgm

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

shmoe

See the identity in Moo Of Doom's post.

soulflyfgm

ok...then what would I take as my induction hypothesis?....plz help..i will apriciate if some one could help me start this proof by induction staring the first steps.. thank you so much

shmoe

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

soulflyfgm

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{r-1} 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

fourier jr

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.

soulflyfgm

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

robert Ihnot

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.