Sum of product of binomials

Question

While working on a combinatorics problem, I found that this result had to be true:

$$\sum_{i=0}^n\frac{(a-b)^i(b-c)^{n-i}}{i!(n-i)!}=\frac{(a-c)^n}{n!}$$

for $a\geq b\geq c$, with $a,b,c\in\Bbb N$.

I tried expanding the binomial powers by using the binomial theorem and then trying to simplify everything, but I didn't succeed. The result reminds me somehow of the product of exponentials property: $e^{a-b}e^{b-c}=e^{a-c}$

Thank you in advance.

Answer 1

Easy peasy....

$\sum_{i=0}^n \frac{(a-b)^i(b-c)^{n-i}}{i!(n-i)!}=\frac{1}{n!}\sum_{i=0}^n \binom{n}{i}(a-b)^i(b-c)^{n-i}=\frac{1}{n!}[(a-b)+(b-c)]^n=\frac{1}{n!}(a-c)^n$

EDIT:sorry I was still typing while the comment has been posted