# Misunderstanding part of pascals identity algebraic proof

1. Aug 12, 2013

### CuriousBanker

I understand the combinatorial proof and the common sense behind why it works but lately I am trying to play around with proofs since I am still new to them. So I understand part of this:

Where I am getting confused is the step where we combine terms. The denominator of the first term is (n-k+1)k!(n-k)! , and the denominator of the second term is k(k-1)!(n-k+1)!

Then on the next line the denominator is k!(n-k+1)!.

How is that a common denominator for those two terms? I see there is a (n-k+1)! in both terms.....but in the second term there is a k(k-1)! and in the first term there is a k!(n-k)!....how do those two terms somehow both reduce to k!?

2. Aug 12, 2013

### CompuChip

It works because of the way the factorial is defined:
$$k! = k \cdot \underbrace{(k - 1) \cdot (k - 2) \cdots 2 \cdot 1}_{{} = (k - 1)!}$$
Similarly, you can show that (n - k + 1) * (n - k)! = (n - k + 1)!.

This is actually the reason for taking the first step: multiplying denominator and numerator by the same number so that the denominators become equal.

3. Aug 14, 2013

### CuriousBanker

Ok, that makes sense. But how does [(n-k+1)n!kn!]/[k!(n-k+1)!] simpylify to [(n+1)n!]/[k!((n+1)-k)!]

Totally lost as to how that step happened

4. Aug 14, 2013

### skiller

For the numerator,
$$(n-k+1)n!kn!$$can you see that this is:
$$[(n-k+1)+k]n!$$ ?

...which equals:
$$(n+1)n!$$
For the denominator, surely you can see that $n-k+1=(n+1)-k$

5. Aug 14, 2013

### CuriousBanker

How? (n−k+1)n!kn! = [(n-k+1)n!]*kn!, whereas [(n−k+1)+k]n! = [(n-k+1)n!]+kn!

Yeah, that part is obvious

6. Aug 14, 2013

### skiller

The numerator is actually $(n-k+1)n!+kn!$ which is $[(n-k+1)+k]n!$ which is $(n+1)n!$