Combinatorial explanation of ${n\choose r}={n-1\choose r-1}+{n-1\choose r}$

Question

${n\choose r}={n-1\choose r-1}+{n-1\choose r}$ (4.1)

Equation（4.1）may be proved analytically or by the following combinatorial argument：Consider a group of $n$ objects, and fix attention on some particular one of these objects-call it object 1. Now there are ${n-1\choose r-1}$groups of size $r$ that contain object 1（since each such group is formed by selecting $r- 1$ from the remaining $n-1$ objects).Also,there are ${n-1\choose r}$groups of size$r$that do not contain object 1. As there is a total of ${n\choose r}$ groups of size $r$,Equation（4.1）follows.

Here's why I feel it is so wrong. Nothing to say about first statement of proof. ${n-1\choose r-1}$ groups of "size $r$ " is in the explanation so I assume it is saying that each group has $r$ object but this is totally wrong since it is $n-1$ choose $r-1$ so it should have size of $r-1$ and why taking ${n-1\choose r-1}$ and saying this group contain the object one. I am sure $n$ is the decision maker of if obj-1 exist or not having $r-1$ does nothing. $n-1$ as I know may or may not have obj-1 because you may have choose the obj-1 and remove it from $n$. ${n-1\choose r}$ so how does changing from $r-1$ to $r$ effects that obj-1 is in that group or not? This explanation is so unnecessary and unclear. I was thinking what size is he talking about. So can anyone say If I am right or wrong? And explain me what actually is being said in human language which is really understandable?

Answer 1

You have $n$ objects. Consider groups of $r$ objects selected from those $n$. They may or may not contain Object $1$. Consider the set containing all objects but Object $1$. It has size $n-1$. So select $r-1$ objects from that set of $n-1$ objects different from Object$1$. You get ${n-1\choose r-1}$ groups of $r-1$ objects all different from Object$1$. Put Object$1$ inside each of them and you'll find the total number of groups of objects of size $r$, containing Object$1$. Now compute the number of groups of objects of size $r$, which do not contain Object $1$. You have a total number of $n-1$ objects (because you don't want Object$1$ to lie among them) and you have to form groups of objects of size $r$, hence that number is ${n-1\choose r}$

Answer 2

You are right and wrong. The first group term does indeed denote choosing $r-1$ objects out of $n-1$, since you are excluding the object you are focusing on (Object $1$). The second term denotes choosing objects with the Object $1$ deliberately excluded. Thus, only $n-1$ objects are there, but we can have to choose an $r$-set out of them. In total: $\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$

Answer 3

The count for ways to select any $r$ from $n$ objects is: of course, $\tbinom{n}{r}$.

We can count this partitioned on whether you include a particular object or not. First we can count ways to select one from one particular object and $r-1$ from the $n-1$ other objects. Secondly we can count ways to select none from one particular object and $r$ from the $n-1$ remaining objects. The sum of these is the count of ways to select $r$ from $n$ objects. $$\binom{n}{r}~=~\binom{1}{1}\binom{n-1}{r-1}+\binom{1}{0}\binom{n-1}{r}$$

Since in each term the first factor equals one, this simplifies.

$$\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$$