$(a+b)^n$ where $a=b=1$

Question

So I have done this question from Boas, Mathematical Methods in Physical sciences, which is

In the expansion of $(a + b)^n$ (see Example 2), let $a = b = 1$, and interpret the terms of the expansion to show that the total number of combinations of n things taken $1, 2, 3, · · · , n$ at a time, is $2^n − 1$

and here is my solution

$2^n=\begin{pmatrix}n\\ 0\end{pmatrix}+\begin{pmatrix}n\\ 1\end{pmatrix}+.....\begin{pmatrix}n\\ r\end{pmatrix}+\begin{pmatrix}n\\ r+1\end{pmatrix}+.....\begin{pmatrix}n\\ n-1\end{pmatrix}+\begin{pmatrix}n\\ n\end{pmatrix}$

as $\begin{pmatrix}n\\ 0\end{pmatrix}=1$

then

$2^n-1=\begin{pmatrix}n\\ 1\end{pmatrix}+.....\begin{pmatrix}n\\ r\end{pmatrix}+\begin{pmatrix}n\\ r+1\end{pmatrix}+.....\begin{pmatrix}n\\ n-1\end{pmatrix}+\begin{pmatrix}n\\ n\end{pmatrix}$

From perivous questions I have done in this book these types of question are generally trying to teach you something, like a concept, but I am having a hard time in seeing what it trying to convey.

What I mean by this is what type of realy life question, physics or maths would this be applied to?

Or am I over thinking it and it was just a question that has no meaning.

edit: The see example 2 refers to $\left(a+b\right)^n=\sum ^n_{r=0}\begin{pmatrix}n\\ r\end{pmatrix}a^{n-r}b^r$

Is it supposed to be $2^n-1$, not $2n-1$ in your problem statement? — Arthur, Aug 30 '18 at 13:11
You already mentioned the combinatorical meaning. Maybe, in statistical mechanics, this equality can be of any use. — Peter, Aug 30 '18 at 13:13
Combinatorial identities can be applied to many things, this is a broad question, but really not new, see e.g. here. — Dietrich Burde, Aug 30 '18 at 13:15
James2018, I found the answers in number-of-subsets-of-a-set-having-r-elements useful. — user376343, Aug 30 '18 at 13:25

score 1 · Accepted Answer · answered Aug 30 '18 at 13:15

What you have is the total number of ways to choose at least one object from a group of $n$ objects.

Let's say you have $4$ different kinds of food on offer at a buffet, but since you had already paid for the buffet, then it would make sense to fill your plate up with at least one item.

So what you have shown above is that there are $2^n - 1 = \sum_{i = 1}^n \binom{n}{i}$ ways to fill your plate up. And in this case of $n = 4$ we have $15$ ways

Let's say it's chicken, Mac n cheese, mashed potatoes, and cake. One way to arrange your plate is just cake, or just chicken. Or maybe instead you want two things, say, chicken and Mac n cheese. Or maybe you want $3$ items. And if you're feeling extra hungry, you could also have the case where you choose all four foods.

score 1 · Answer 2 · answered Aug 30 '18 at 13:18

It is useful in combinatorics, and on any field that involves probability.

As an example, you can consider the amount of surjections from $[[1,n]]$ to $ \{1,2 \}$. There are 2 applications that are not surjective among the $2^n$ possibilities.

The $2^n$ result can be found by summing all of the possibilities of applications: $\sum_{k=0}^{n} C_k^n = 2^n$.

I think they try to convey the fact that it's useful for probability fields, and that you can count with $C_k^n$ (see example above).

$(a+b)^n$ where $a=b=1$

2 Answers2