Let $A$ be a set with $n$ elements. The involution $S \mapsto A\setminus S$ shows that $\binom{n}{k} = \binom{n}{n-k}$ for $0 \leqslant k \leqslant n$, so we can restrict our attention to $k \leqslant m := \bigl\lfloor \frac{n}{2}\bigr\rfloor$.
Suppose we want to select a committee of $m$ people, among which $k$ shall represent the committee to the outside world from a group of $n$ people. We can
- Select the $m$ members of the committee first, and then choose the $k$ representatives from the committee members afterwards. That is possible in $\binom{n}{m}\cdot \binom{m}{k}$ ways.
- First choose the $k$ representatives from the whole group, and then select the remaining $m-k$ committee members from the remaining $n-k$ group members. That is possible in $\binom{n}{k}\cdot \binom{n-k}{m-k}$ ways.
Using the symmetry of the binomial coefficients, it follows that
$$\binom{n}{m} = \frac{\binom{n-k}{m-k}}{\binom{m}{m-k}}\cdot \binom{n}{k}.\tag{1}$$
Now we note that for $0 \leqslant r \leqslant s \leqslant t$ we have
$$\binom{s}{r} \leqslant \binom{t}{r},\tag{2}$$
with equality if and only if $s = t$ or $r = 0$.
Since $n-k \geqslant n-m \geqslant m$, using the inequality $(2)$ in $(1)$ shows the desired
$$\binom{n}{m} \geqslant \binom{n}{k},$$
with equality if and only if $k = m$.
You are misinterpreting the problem. When asked to find the sum, say for example $2$, you have to count $(0, 1), (1, 0), (0, 2), (2, 0), (0, 0), (1, 1)$. It is nowhere given that the order of $x_i$ doesn't matter.
Now note that we only deal with $f(a, b)$ such that $a = b$, so I will not use $b$ now, will only use $a$.
You correctly simplified the expression $$\sum_{\displaystyle\sum_{i=1}^{a}x_i\leq a}\dfrac{a!}{x_1!\cdot x_2!\cdot\text{$\dots$}\cdot x_a!\cdot (a-x_1-x_2-\dots-x_a)!}$$
Now look at the denominator. Do you notice something?
It's just all the $a + 1$ sequences with sum $a$.
So, expression equals $$\sum_{x_1 + x_2 + \cdots + x_{a + 1} = a} \binom{a}{x_1, x_2, \cdots, x_{a + 1}}$$
Can you identify this as the multinomial coefficient?
The expression equals $$\sum_{x_1 + x_2 + \cdots + x_{a + 1} = a} \binom{a}{x_1, x_2, \cdots, x_{a + 1}} 1 ^ {x_1} 1 ^ {x_2}\cdots 1^{x_{a + 1}}$$
This equals $$(1 + 1 + \cdots + 1) ^ {a} = (a + 1) ^ a$$
So, we are required to find $$\prod_{x = 2}^{100}{(x + 1)^x}$$
The number of power of $2$ is greater than the number of power of $5$ in this expression, and since we are interested in $$\operatorname{min}(v_2, v_5)$$ we count the power of $5$.
We count with a method similar to Legendre's Formula.
First, count the power of $5$. This is $$4 + 9 + 14 + \cdots + 99 = 1030$$
Then, count the power of $25$. This is $$24 + 49 + 74 + 99 = 246$$
Total: $$246 + 1030 = 1276$$
Best Answer
Let’s take a look:
This is part of a proof by induction that
$$\sum_{k=0}^n\binom{m+k}m=\binom{m+1+n}{m+1}\tag{1}$$
for all $n\in\Bbb N$. Let $P(n)$ be the specific assertion in $(1)$, that the identity is true for the particular value $n$. Then the calculation above is the induction step: the induction hypothesis is that $P(n-1)$ is true, and we’re showing that this implies that $P(n)$ is true. $P(n-1)$, written out in full, is that
$$\sum_{k=0}^{n-1}\binom{m+k}m=\binom{m+1+(n-1)}{m+1}=\binom{m+n}{m+1}\;;\tag{2}$$
we’ll use this in a moment.
The first equality in the calculation is simply splitting the $k=n$ term off from the rest of the summation. The second step uses $(2)$, the induction hypothesis, to replace $\sum_{k=0}^{n-1}\binom{m+k}m$ by its value $\binom{m+n}{m+1}$. And the third and last equality is just Pascal’s identity, which in its general form states that
$$\binom{n+1}k=\binom{n}k+\binom{n}{k-1}\;.\tag{3}$$
This can be proved either combinatorially or algebraically, as you set out to do with the special case needed for the calculation here. If you’re going to prove it algebraically, though, it’s easier to prove the general form $(3)$ and then substitute into it to get the special case that you need.