# Thread: Proofs im having trouble with

1. ## Proofs im having trouble with

Prove Pascals Formula by using the fact that nCk = n!/(k!(n-k)!)

Prove that k*(nCk) = n*(n-1)C(k-1) by using the same fact from above.

Use the lattice walk idea to prove that nCk = nC(n-k).

Use the lattice walk idea to prove nC0 + nC1 + ... + nCn = 2^n

If anyone can show any of these, I would greatly appreciate it. Thanks!

2. Originally Posted by jzellt
Prove Pascals Formula by using the fact that nCk = n!/(k!(n-k)!)

Prove that k*(nCk) = n*(n-1)C(k-1) by using the same fact from above.

Use the lattice walk idea to prove that nCk = nC(n-k).

Use the lattice walk idea to prove nC0 + nC1 + ... + nCn = 2^n

If anyone can show any of these, I would greatly appreciate it. Thanks!

I use $\binom{n}{k}$ instead of $nCk$ , so:

$\displaystyle{k\binom{n}{k}=k\frac{n!}{k!(n-k)!}=\frac{n!}{(k-1)!(n-k)!}$ , since $\displaystyle{m(m-1)!=m!\Longleftrightarrow \frac{m}{m!}=\frac{1}{(m-1)!}}$

Now show that the above equals $\displaystyle{n\binom{n-1}{k-1}}$

I don't know what the path walk idea is, but the last equality follows at once

from Newton's Binom: $\displaystyle{(a+b)^n=\sum\limits^n_{k=0}\binom{n} {k}a^{n-k}b^k}$

Tonio