In regards to the arithmetic triangle:
1
1 1
1 2 1 etc
we know that it is a way to calculate the binomial coefficients or number of choices $C(n,r)$.
So I understand that there is only $1$ way to choose a set of size $0$ from an empty set and that is what the $1$ at the top of the triangle shows.
Also there is $1$ way to choose a set of size $0$ from any set hence each row starts with $1$. Each row is built on the previous one and I am confused with the logic for $2$, in the center of the third line.
Basically that $2$ means that there are $2$ ways to pick a single person from a pair, but this $2$ is the sum of the ones in the first row i.e. $1$ way to choose a set of size $0$ and $1$ way to choose one person from a set of one.
But summing these two with this meaning does not make sense to me i.e. that the number of ways to pick a single person from a pair is the number of ways to pick $0$ persons from an empty set plus the number of ways to pick $1$ person from a pair.
What am I misunderstanding here?
It might be more helpful to think about the general case first: $\binom{n}{r} + \binom{n}{r + 1} = \binom{n+1}{r+1}$. (This is the combinatorial formula behind the condition that a number in the triangle is the sum of the two numbers above it.)
To explain this formula, imagine you have people numbered $1, 2, \dots, n + 1$, and that you want to choose a subset of $r + 1$ of them. The key idea is that you will include person $n + 1$ or you won’t. If you don’t include person $n + 1$, then you’re choosing a $(r + 1)$-subset from $n$ people, i.e. $\binom{n}{r + 1}$. If you do include person $n + 1$, then you have to choose $r$ more people from $n$ people, i.e. $\binom{n}{r}$. The “either/or” setup implies that the total number of subsets $\binom{n+1}{r+1}$ is the sum of these two binomial coefficients.
Now as for your special case: $\binom{2}{1} = \binom{1}{0} + \binom{1}{1}$, the people are named $1$ and $2$, and we want to choose just 1 of them. Your subset either contains person 2 or it doesn’t. If it doesn’t, we get the $\binom{1}{1} = 1$ term as we are forced to take person 1. If the subset does contain person 2, then we choose $0$ people from a set of size 1, i.e. the $\binom{1}{0}$ term.