Let $\mathbb{A}=\{a_1,\ a_2,\ \cdots,\ a_{100}\}$ be a set containing $100$ real numbers, $\mathbb{B}=\{b_1,\ b_2,\ \cdots,\ b_{50}\}$ be a set containing $50$ real numbers, and $\mathcal{F}$ be a mapping from $\mathbb{A}$ to $\mathbb{B}$. Find the number of possible $\mathcal{F}$ if $\mathcal{F}(a_1) \le \mathcal{F}(a_2)\le\cdots\mathcal{F}(a_1)$, and for every $b_i\in\mathbb{B}$, there exists an element $a_i\in\mathbb{A}$ such that the $\mathcal{F}(a_i)=b_i$.
Without loss of generality, let's assume $b_1 < b_2 < \cdots < b_{50}$. Then the answer is the positive integer solutions to the following equation $$x_1 + x_2 + \cdots + x_{50} = 100$$
where $x_i$ denotes the number of $a_j\in\mathbb{A}$ such that $\mathcal{F}(a_j)=b_i$. Hence the answer is $$\binom{100-1}{50-1}=\boxed{\binom{99}{49}}$$