Chapter 2.6, Problem 20E

(a)

To determine

To prove: $\left(\begin{array}{c}n\\ r\end{array}\right)=\left(\begin{array}{c}n\\ n-r\end{array}\right)$ using a combinatorial argument.

Explanation of Solution

Given information: Use a combinatorial argument.

Proof:

Let n and r be any natural number, $r\le n$ .

  $\left(\begin{array}{c}n\\ r\end{array}\right)=$ number of ways you can choose r objects out of n objects

If you choose r objects, n − r objects are left, so this is equivalent to choosing n − r objects out of n objects.

Number of ways you can choose n − r objects out of n objects $=\left(\begin{array}{c}n\\ n-r\end{array}\right)$

Therefore,

  $\left(\begin{array}{c}n\\ r\end{array}\right)=\left(\begin{array}{c}n\\ n-r\end{array}\right)$

(b)

To determine

To prove: $\left(\begin{array}{c}n\\ r\end{array}\right)=\left(\begin{array}{c}n-1\\ r\end{array}\right)+\left(\begin{array}{c}n-1\\ r-1\end{array}\right)$ using a combinatorial argument.

Explanation of Solution

Given information: using a combinatorial argument.

Proof:

Let n and r be any natural number, $r\le n$ . Let A be a set of n different elements, let a be any chosen element of A.

There are more ways of choosing r elements of A.

First, out of all elements choose r elements.

  $\left(\begin{array}{c}n\\ r\end{array}\right)=$ number of ways you can choose r objects out of n objects

Second case, some chosen subsets contain a and some don’t.

Number of sets that don’t contain a and have r elements is $\left(\begin{array}{c}n-1\\ r\end{array}\right),$ choose r elements of $A-\left\{a\right\},n-1$ elements.

Number of sets that contain a is $\left(\begin{array}{c}n-1\\ r-1\end{array}\right)$ , besides a choose r − 1 more elements of $A-\left\{a\right\},n-1$ element.

This is one more way of chosen r elements out of a set of n elements.

Therefore,

  $\left(\begin{array}{c}n\\ r\end{array}\right)=\left(\begin{array}{c}n-1\\ r\end{array}\right)+\left(\begin{array}{c}n-1\\ r-1\end{array}\right)$

(c)

To determine

To prove: $\left(\begin{array}{c}n-1\\ r\end{array}\right)+\left(\begin{array}{c}n-1\\ r-1\end{array}\right)=\left(\begin{array}{c}n\\ r\end{array}\right)$ algebraically.

Explanation of Solution

Given information:algebraically.

Proof:

The proof form (b) but algebraic

  $\begin{array}{c}\left(\begin{array}{c}n-1\\ r\end{array}\right)+\left(\begin{array}{c}n-1\\ r-1\end{array}\right)\stackrel{\text{the definition}}{=}\frac{\left(n-1\right)!}{r!\left(n-1-r\right)!}+\frac{\left(n-1\right)!}{\left(r-1\right)!\left(n-r\right)!}\\ \stackrel{\text{expand}}{=}\frac{\left(n-1\right)!}{r\left(r-1\right)!\left(n-1-r\right)!}+\frac{\left(n-1\right)!}{\left(r-1\right)!\left(n-r\right)\left(n-r-1\right)!}\\ \stackrel{\text{combine}}{=}\frac{\left(n-1\right)!}{\left(r-1\right)!\left(n-1-r\right)!}\left(\frac{1}{r}+\frac{1}{n-r}\right)\\ \stackrel{\text{simplify}}{=}\frac{\left(n-1\right)!}{\left(r-1\right)!\left(n-1-r\right)!}\left(\frac{n-r+r}{r\left(n-r\right)}\right)\\ =\frac{\left(n-1\right)!}{\left(r-1\right)!\left(n-1-r\right)!}\left(\frac{n}{r\left(n-r\right)}\right)\\ =\frac{n!}{\left(r\right)!\left(n-r\right)!}\\ \left(\begin{array}{c}n-1\\ r\end{array}\right)+\left(\begin{array}{c}n-1\\ r-1\end{array}\right)=\left(\begin{array}{c}n\\ r\end{array}\right)\end{array}$

(d)

To determine

To prove: ${\left(a+b\right)}^{n+1}={\displaystyle \sum _{r=0}^{n+1}\left(\begin{array}{c}n+1\\ r\end{array}\right)a^r{b}^{n+1-r}}$ by induction.

Explanation of Solution

Proof:

Let aandb any real numbers.

1. Basic step:

  $n=1\Rightarrow {\left(a+b\right)}^1=a+b=\left(\begin{array}{c}1\\ 0\end{array}\right)a+\left(\begin{array}{c}1\\ 1\end{array}\right)b$

2. Inductive step:

Assume, that for some natural number n , ${\left(a+b\right)}^n={\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)a^r{b}^{n-r}}$ .

Then, for n + 1:

  ${\left(a+b\right)}^{n+1}\stackrel{\text{isolate the last element}}{=}{\left(a+b\right)}^n\left(a+b\right)$

  $\stackrel{\text{assumption}}{=}\left({\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)a^r{b}^{n-r}}\right)\left(a+b\right)$

  $\stackrel{\text{combine}}{=}{\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)a^{r+1}{b}^{n-r}}+{\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)a^r{b}^{n-r+1}}$

  $\text{ =}{\displaystyle \sum _{r=1}^{n+1}\left(\begin{array}{c}n\\ r-1\end{array}\right)a^r{b}^{n-\left(r-1\right)}}+{\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)a^r{b}^{n-r+1}}$

  $\stackrel{\text{isolate the first element}}{=}\left(\begin{array}{c}n\\ 0\end{array}\right)a^0{b}^{n+1}+{\displaystyle \sum _{r=1}^{n+1}\left(\begin{array}{c}n\\ r-1\end{array}\right)a^r{b}^{n+1-r}}+{\displaystyle \sum _{r=0}^n\left(\begin{array}{c}n\\ r\end{array}\right)a^r{b}^{n+1-r}}$

  $\stackrel{\text{combine the sums}}{=}\left(\begin{array}{c}n\\ 0\end{array}\right)a^0{b}^{n+1}+{\displaystyle \sum _{r=1}^{n+1}\left(\left(\begin{array}{c}n\\ r-1\end{array}\right)a^r{b}^{n+1-r}+\left(\begin{array}{c}n\\ r\end{array}\right)a^r{b}^{n+1-r}\right)}$

  $\stackrel{}{=}\left(\begin{array}{c}n\\ 0\end{array}\right)a^0{b}^{n+1}+{\displaystyle \sum _{r=1}^{n+1}\left(\begin{array}{c}n+1\\ r\end{array}\right)a^r{b}^{n+1-r}}$

  ${\left(a+b\right)}^{n+1}={\displaystyle \sum _{r=0}^{n+1}\left(\begin{array}{c}n+1\\ r\end{array}\right)a^r{b}^{n+1-r}}$