Link between thetrahedral numbers and combinatorics The triangular numbers are $1, 3=2+1, 6=3+2+1$ and the $n$-th triangular number is
$$\binom{n+1}{2}=\frac{n(n+1)}{2}=n+(n-1)+\ldots+2+1.$$
There is a neat explation that the n-te triangular number is  $\binom{n+1}{2}$: Consider $n+1$ people. Then there are $\binom{n+1}{2}$ pairs. But you could also count the pairs differently (see here): You can match the first person with $n$ different persons, the second person can be paired with $n-1$ different persons, the third with $n-2$, $\ldots$. Hence $\binom{n+1}{2}=n+(n-1)+(n-2)+\ldots$.
Now the tetrahedral numbers $1, 4=1+3, 10=1+3+6,\ldots$ can be written as $\binom{n+2}{3}$. Is there also an combinatorical explaination as for triangular numbers? I just found this explanation but I can't see clearly why $\binom{n+2}{3}$ should be the same as $1+3+6+\ldots+\frac{n(n+1)}{2}$.
 A: Your last link says

The numbers you're seeing are binomial coefficients. To go from $(0,0)$ to $(3,n)$ you have to take $3+n$ steps: $3$ right and $n$ up. You have to choose which $3$ of the $3+n$ steps that are going to go to the right, and the number of ways to choose is $\binom{3+n}{3}$

Your question is why $\binom{n+3}{3} =\sum\limits_{m=0}^n  \binom{m+2}{2}$
In the context of the link, this says the number of ways of going $3$ right and $n$ up is equal to the number of ways of going $2$ right and $m$ up in any pattern, followed immediately by $1$ right and then by $n-m$ up, summed over all the possible $m$. In other words all the possible ways, measured by how many steps up before the last step right.
A: I found an analogous explaintion using people.
Let's assume we want to calculate $\binom{5}{3}$, i.e. the number of subsets of order 3 with 5 elements / people {A,B,C,D,E}.

*

*Subgroups of size 3 with A are all possible pairs using {B,C,D,E} and
A itself, i.e. $\binom{4}{2}$.


*Subsets of size 3 with B (and without A, because we counted all
subsets with A already) are all  possible pairs using {C,D,E} and B
itself, i.e. $\binom{3}{2}$.


*Subsets of size 2 with C (and without A and B, because we counted
these subsets already) are all possible pairs using {D,E} and C
itself, i.e. $\binom{2}{2}$.
Hence all possible subsets are $\binom{4}{2}+\binom{3}{2}+\binom{2}{2}=\binom{5}{3}$ and in general
$$\binom{n+2}{3}=\binom{2}{2}+\binom{3}{2}+\binom{4}{2}+\ldots +\binom{n+1}{2}.$$
