Proving equality of factorial expressions

Question

I was solving some combination problems and noticed that some of them can have two or more different ways to solve.
Look at these examples :

In how many ways can we wear a shoes and b shirts together:
1-One of the ways is just multiplying a and b, which means the answer is ab.
2-Another way is calculating number of ways that we can pick two objects from all objects, which is $\binom{a+b}{2}$. But it is possible that we've picked two shirts or two shoes, As a result we should subtract the number of ways that we can pick up two shirts and the number of ways that we can pick two shoes from $\binom{a+b}{2}$. Which means the answer is: $\binom{a+b}{2}$ - $\binom{a}{2}$ - $\binom{b}{2}$

Another question is this : set S equals to {1,2,3,4,.....,n}. How many subsets of S have m members and their smallest member is smaller than m? Again we have two ways to solve this question:
1-First we calculate number of m-member subsets of S which is $\binom{n}{m}$. Then we should subtract number of sets that their smallest member is not smaller than m. which means we should choose subset members from m to n, that is $\binom{n-m+1}{m}$. Thus, the answer is $\binom{n}{m}$ - $\binom{n-m+1}{m}$.
2-We know that if subset's smallest member is 1, We have $\binom{n-1}{m-1}$ ways to choose subset.(Because we've already chosen one of members and we can choose the rest from 2 to n, which is n-1). But the smallest member can also be 2. It means we have $\binom{n-2}{m-1}$ options if smallest value is 2. smallest value can also be 3, 4, ....., m-1. That means the answer of this question is :
$\binom{n-1}{m-1}$ + $\binom{n-2}{m-1}$ + $\binom{n-3}{m-1}$ + .... + $\binom{n-m+1}{m-1}$

It is obvious that ab always is equal to $\binom{a+b}{2}$ - $\binom{a}{2}$ - $\binom{b}{2}$ if a,b∈N, and $\binom{n-1}{m-1}$ + $\binom{n-2}{m-1}$ + $\binom{n-3}{m-1}$ + .... + $\binom{n-m+1}{m-1}$ would always equal $\binom{n}{m}$ - $\binom{n-m+1}{m}$ if m,n∈N and n>m.(Because they produce answers for same question.)
Now I'm interested in proving these equalities algebraically. when I replace combinations with formula $\frac{n!}{r!(n-r)!}$, It seems really hard to prove the equality. Please help me do it.
Is it possible that some equality of factorial expression exists that has no algebra proof? and the only way to prove that is combinatorial proof?

Answer 1

$$ {n\choose m}-{n-m+1\choose m}=\left[{n\choose m}-{n-1\choose m}\right]+\left[{n-1\choose m}-{n-2\choose m}\right]+\dotsb $$

$$ +\left[{n-m+2\choose m}-{n-m+1\choose m}\right]={n-1\choose m-1}+{n-2\choose m-1}+\dotsb+{n-m+1\choose m-1} $$