Proving P(n) for nCr Using Induction

[soulflyfgm]

i have this so far..please tell me if this is right..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( here i don't know wat r is..im guessing r has to be 0 because of the 0<= r<= n)

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

 

[matt grime]

0C0 is 1, not 0. apart from that your idea is correct, though you also need to note that nC0 and nCn are integers, otherwise the identity you've written down

(n+1)Cr = nCr +nC(r-1)

is no help if r=n+1 or 0 then one of the terms on the left is undefined (by you).