Can n choose k be proven as a natural number without using induction?

[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

$$\binom {n}{r} = \frac {n \times (n-1) \times \cdot \cdot \cdot \times (n-r+1)}{r \times (r-1) \cdot \cdot \cdot \times 1}$$

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]

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.

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]

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

 

[ComputerGeek]

A third approach is to observe that in
$$\binom {n}{r} = \frac {n \times (n-1) \times \cdot \cdot \cdot \times (n-r+1)}{r \times (r-1) \cdot \cdot \cdot \times 1}$$
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.

I thought that
$$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$

 

[shmoe]

I thought that
$$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$

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

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

 

[soulflyfgm]

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

 

[shmoe]

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

See the identity in Moo Of Doom's post.

 

[soulflyfgm]

See the identity in Moo Of Doom's post.

ok...then what would I take as my induction hypothesis?...please 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 please 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.

 

[shmoe]

A third approach is to observe that in
$$\binom {n}{r} = \frac {n \times (n-1) \times \cdot \cdot \cdot \times (n-r+1)}{r \times (r-1) \cdot \cdot \cdot \times 1}$$
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.

Unfortunately you can't pair them up easily to cancel terms as you may have terms in the numerator pulling "double duty" and being the only term in the numerator divisble by multiple terms in the denominator. Consider 7C3=(7*6*5)/(1*2*3), the 2 and the 3 both go to the 6.

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{r-1} doesn't hold for all values of r.

 

[soulflyfgm]

proof

Unfortunately you can't pair them up easily to cancel terms as you may have terms in the numerator pulling "double duty" and being the only term in the numerator divisble by multiple terms in the denominator. Consider 7C3=(7*6*5)/(1*2*3), the 2 and the 3 both go to the 6.

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{r-1} doesn't hold for all values of r.

do u think this prove is correct

Let P(n) be the statement that for any n in the natural numbers N, nCr is an element of N for every r with 0<= r<= n
nCr = n!/(r!(n-r)!)
0Cr = o!/(r!(0-r)!) = 0
so P(0)E(belongs) in N (natural numbers)
Suppose P(n) is a natural number of any N
then since {n+1}Cr = nCr + nC{r-1} is true ( already proved it algebraically and i will add it to this part) so 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 review this prove and tell me if its right? thank you so much for ur help