choose $k$ numbers $a_1, a_2, …, a_k$ from the set $\{1,2,…,m\}$ such that $a_1 \le a_2 \le a_3 \le ⋯ \le a_k$ Assume that $k$ and $m$ are positive integers such that $k \le m$. Determine the number of ways in which one can choose $k$ numbers $a_1, a_2, …, a_k$ from the set $\{1,2,…,m\}$ such that $a_1 \le a_2 \le a_3 \le ⋯ \le a_k$.
I think I can use an idea with the bijection principle to solve this, I just do not see how. Here is the idea:
assume $b_i$ is the difference between two consecutive integers from our chosen sequence, $b_1=a_1, b_2=a_2-a_1, \dots b_k=a_k-a_{k-1}, b_{k+1}=m$, when we find the sum of all of the $b_i$, we get the number $m$.
How would this help me solve this problem? Would I even be able to prove that there is a bijection here when I have $k$ numbers in the sequence and $k+1$ $b_i$ terms?
edit:
I also noticed that for every $m$ and $k$ there are $m+k-1 \choose k$ ways to choose $k$ numbers from the set. So for example if $m=3$ and $k=2$ would the total number of ways we can choose k numbers from the set be $3+2-1 \choose 2$? In general the answer to this question would be $m+k-1 \choose k$?
****Solution:
We need to determine $|S|$ where
$$S=\{(a_1, \dots, a_k) \in \{1, 2, \dots, m\}: a_1 \le a_2 \le \cdots \le a_k\}$$
Let us define the set $T$ in the following way:
$$T=\{(b_1, \dots , b_{k+1}) \in \mathbb{N}_0^{k+1}:b_1+b_2+\cdots +b_{k+1}=m-1\}$$
Let $f:S \to T$ be the function defined in the following way:
$$f(a_1,a_2,\dots,a_k)=(a_1-1,a_2-a_1,a_3-a_2,\dots,a_k-a_{k-1},m-a_k)$$
Now it is obvious that $f$ is a bijection, therefore $|S|=|T|$. Now we can calculate $|T|$ easily because this is the same as the number of ways to distribute $m-1$ objects to $k+1$ people. This is similar to arranging $m-1$ objects with $k$ dividers. Hence, $|S| =$ $m+k-1 \choose k$.
 A: With the understanding that:


*

*the condition imposed is correctly $a_{\,1}  \leqslant a_{\,2}  \leqslant  \cdots  \leqslant a_{\,k} $ and not $a_{\,1}  < a_{\,2}  <  \cdots  < a_{\,k} $;

*the "set" $\left\{ {1,3,2} \right\}$ is equivalent to $\left\{ {1,2,3} \right\}$ 
and the "multiset" $\left\{ {2,1,2} \right\}$ is equivalent to $\left\{ {1,2,2} \right\}$,
so that, standing the condition above, we cannot speak in terms of "sets";

*that the problem should then correctly be reworded as
what is the number of non-decreasing words, of length $k$, from alphabet $\left\{ {1,2, \dots , m} \right\}$.  


Then your method is "nearly correct" in the sense that, while it is a good approach to consider the delta of the terms, it is not to be taken as you defined it, because summing the $b_j$ from $1$ to $k$ you obtain $a_k$, and summing from $1$ to $m+1$ you get $a_k+m$.
Instead keep the first part of your definition
$$
b_{\,1}  = a_{\,1}  - 0,\quad b_{\,2}  = a_{\,2}  - a_{\,1} ,\quad  \cdots \;,\quad b_{\,k}  = a_{\,k}  - a_{\,k - 1} 
$$
  ------ amended part ------
then you have:
$$
\left\{ \begin{gathered}
  a_{\,1}  = b_{\,1}  \hfill \\
  0 \leqslant b_{\,j}  \leqslant a_{\,k}  \leqslant m\quad \left| {2 \leqslant j} \right. \hfill \\
  \sum\limits_{2\, \leqslant \,j\, \leqslant \,k} {b_{\,j} }  = a_{\,k}  - a_{\,1}  \hfill \\ 
\end{gathered}  \right.
$$
so that the number of ways in which you can choose the $b_2 \cdots b_k$ values
corresponds to the number of weak k-compositions of $a_k-a_1$ which
is readily attainable to be:
$$
\left( \begin{gathered}
  a_{\,k}  - a_{\,1}  + k - 2 \\ 
  k - 2 \\ 
\end{gathered}  \right)
$$
and which summed for $a_1$ from $1$ to $a_k$ gives
$$
\sum\limits_{1\, \leqslant \,a_{\,1} \, \leqslant \,a_{\,k} } {\left( \begin{gathered}
  a_{\,k}  - a_{\,1}  + k - 2 \\ 
  k - 2 \\ 
\end{gathered}  \right)}  = \left( \begin{gathered}
  a_{\,k}  + k - 2 \\ 
  k - 1 \\ 
\end{gathered}  \right) = \left( \begin{gathered}
  a_{\,k}  + k - 1 \\ 
  k - 1 \\ 
\end{gathered}  \right) - \left( \begin{gathered}
  a_{\,k}  + k - 2 \\ 
  k - 2 \\ 
\end{gathered}  \right)
$$
where the last equality means that we get the same result if we allow the compositions to start also from $0$
and deduct those that actually start with a $0$.  
Then summing for $a_k$ for $1$ to $m$, we arrive at:
$$
N = \sum\limits_{1\, \leqslant \,a_{\,k} \, \leqslant \,m} {\left( \begin{gathered}
  a_{\,k}  + k - 2 \\ 
  k - 1 \\ 
\end{gathered}  \right)}  = \left( \begin{gathered}
  m + k - 1 \\ 
  m - 1 \\ 
\end{gathered}  \right) = \left( \begin{gathered}
  m + k - 1 \\ 
  k \\ 
\end{gathered}  \right)
$$
