Here's a part of a past question from an entrance examination which leads me to some
$$ A_{m,n} = \{ (\alpha_1, \alpha_2, \dotsc, \alpha_m) : 1 \leq \alpha_1 \leq \alpha_2 \leq \dotsb \leq \alpha_m \leq n \} $$ given that $\alpha_i \in \mathbb{Z}$ for all $i$, $$ B_{m,n} = \{ (\alpha_1, \alpha_2, \dotsc, \alpha_m) : \alpha_1 + \alpha_2 + \dotsb + \alpha_m = n \} $$ given that $\alpha_i \geq 0$ and $\alpha_i \in \mathbb{Z}$ for all $i$, $$ C_{m,n} = \{ (\alpha_1, \alpha_2, \dotsc, \alpha_m) : 1 \leq \alpha_1 < \alpha_2 < \dotsb < \alpha_m \leq n \} $$ given that $\alpha_i \in \mathbb{Z}$ for all $i$.
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Are the sets $A,B,C$ bijective in any way? I am not sure of that but the question asked us to find the number of elements in $A_{m,n}$ and the solution says that since $A,B$ form a bijection, they have the same number of elements. Thus they just count the number in any one of $A,B$ and does for the other.
Please explain me how $A$ and $B$ forms a bijection. Also, if $C$ forms a bijection, do explain me that as well.
What I think is that they don't form a bijection directly. But when the number of elements are altered, we can make them form a bijection. Like, $A_{x,y}$ and $B_{r,s}$ might form a bijection where $x$ is not necessarily equal to $r$ and $y$ to $s$. In that case, we can create a bijection using the the idea of a path travelled from the origin to $(x,y)$ in the coordinate plane such that there's only rightward and upward movements.
Hint: $$ 1\le \alpha_1\le \alpha_2\le \dots \le \alpha_m\le n $$ if and only if $$ 1\le \alpha_1<\alpha_2+1< \alpha_3+2<\dots < \alpha_m+m-1\le n+m-1 $$ if and only if $$ (\alpha_1-1) + (\alpha_2-\alpha_1)+\dots+(\alpha_m-\alpha_{m-1})+(n-\alpha_m)=n-1 $$ and all the summands are nonnegative.