Intuition behind sums of sums of whole numbers

I am still not able to comment on this site, so I have to write this as an answer.

Look at the number of ways one is able to choose $2$ balls from a set of $n+1$ numbered balls.

If you chose the ball numbered one, you can choose the second ball in $n$ ways. Now, if you chose ball numbered two as the first ball, then your second ball can be chosen in $n-1$ number ways and so on. The ways of choosing the 2 balls is just $n+n-1+\cdots+1$.

Now, look at the ways of choosing 3 balls from a set of $n+2$ numbered balls. If the first ball you chose is ball number one, then the other twos balls can be chosen in $n+n-1+\cdots+1$ ways, from our last paragraph. Now, if the first ball you chose was ball number two, then the other two could be chosen in $n-1+\cdots+1$ ways and so on.

I hope you see where I'm going with this.

We can write the sum $N_2(n)$ as \begin{align*} N_2(n)&=\sum_{1\leq i}^{n}\sum_{1\leq j\leq i}j =\sum_{1\leq j\leq i\leq n}j =\sum_{1\leq j\leq i\leq n}\sum_{k=1}^j1\\ &=\sum_{\color{blue}{1\leq k\leq j\leq i\leq n}}1\\ &=\binom{n+2}{3} \end{align*}

In general we can write for $k\geq 1$: \begin{align*} N_k(n)&=\sum_{\color{blue}{1\leq j_1\leq j_2\leq \cdots\leq j_{k+1}\leq n}}1\tag{1}\\ &=\binom{n+k}{k+1} \end{align*}

In (1) we observe the index range contains all ordered $k+1$-tuples with elements from $\{1,2,\ldots,n\}$ with repetition. This number is given by the binomial coefficient $\binom{n+k}{k+1}=\binom{n+k}{n-1}$.