Counting Principles, Permutation and Combination
Author:
Hemanga
Last Updated:
10년 전
License:
LaTeX Project Public License 1.3c
Abstract:
My M.Sc project
\begin
Discover why over 20 million people worldwide trust Overleaf with their work.
My M.Sc project
\begin
Discover why over 20 million people worldwide trust Overleaf with their work.
\documentclass{beamer}
\mode<presentation>{
\usetheme{Madrid}
%\usecolortheme{beaver}
}
\usepackage[utf8]{inputenc}
\usepackage{default}
\usepackage[portuguese]{babel}
\usepackage{pgfplots}
\pgfplotsset{/pgf/number format/use comma,compat=newest}
\usepackage{color}
\usepackage{amsmath,amsfonts,amssymb}
\usepackage{hyperref}
\usepackage{tikz}
\usepackage{enumerate}
\usebackgroundtemplate{%
\tikz\node[opacity=0.1] {\includegraphics[height=\paperheight,width=\paperwidth]{tu.jpg}};}
\title{COUNTING PRINCIPLES, PERMUTATION AND COMBINATION}
\author{ \bf HEMANGA DUTTA}
\institute[]{\includegraphics[height=1in]{tu.jpg}{\vspace{.3cm}}\\ \bf M.Sc IN MATHEMATICS{\vspace{.3cm}} \\ ROLL NO. : MSM14039{\vspace{.3cm}}\\ {\bf TEZPUR UNIVERSITY}}
\date{\hspace{8cm} }
\begin{document}
\begin{frame}
\maketitle
\end{frame}
%\begin{frame}
%\frametitle{Sumário}
%\tableofcontents
%\end{frame}
\section{}
\begin{frame}
\frametitle{OUTLINE}
In this presentation, we are going to discuss about,
\begin{enumerate}
\item {\bf Four Basic Counting Principles}
\begin{enumerate}
\item Addition
\item Multiplication
\item Subtraction
\item Division{\vspace{.5cm}}
\end{enumerate}
\item {\bf Permutation}
\begin{enumerate}
\item Permutation of Multiset{\vspace{.5cm}}
\end{enumerate}
\item {\bf Combination}
\begin{enumerate}
\item Combination of multiset
\end{enumerate}
\end{enumerate}
\end{frame}
\section{}
\begin{frame}
\frametitle{Four Basic Counting Principles}
There are {\it \bf FOUR} basics counting principle .they are,{\vspace{.5cm}}
\begin{enumerate}
\item {\Large \bf Addition} {\vspace{.5cm}}
\item {\Large \bf Multiplication}{\vspace{.5cm}}
\item {\Large \bf Subtraction}{\vspace{.5cm}}
\item {\Large \bf Division }
\end{enumerate}
\end{frame}
\section{}
\begin{frame}
\frametitle{Four Basic Counting Principles}
{ \Large {\bf ADDITION PRINCIPLE :}}
{ \it{ Suppose that a set $S$ is partitioned into parts $S_1$,$S_2$,\dots ,$S_m$. The number of of objects in S can be determined by finding the number of objects in each parts, and adding the numbers so obtaine:}}
$$|S| = |S_1|+|S_2|+\dots +|S_m|$$
\end{frame}
\section{}
\begin{frame}
\frametitle{Four Basic Counting Principles}
{ \Large {\bf ADDITION PRINCIPLE :}}
{ \it{ Suppose that a set $S$ is partitioned into parts $S_1$,$S_2$,\dots ,$S_m$. The number of of objects in S can be determined by finding the number of objects in each parts, and adding the numbers so obtaine:}}
$$|S| = |S_1|+|S_2|+\dots +|S_m|$$
{ {\bf Example : }A student wishes to take either a mathematics course or a biology course, but not both. If there are 4 mathematics courses and 3 biology courses for which the student has the necessary prerequisites, then the student can choose a course to take in 4+3=7 ways.}\\
\end{frame}
\section{}
\begin{frame}
\frametitle{Four Basic Counting Principles}
{\Large \bf MULTIPLICATION PRINCIPLE :}
{\it Let $S$ be a set of ordered pair $ (a,b) $ of objects, where the first object $a$ comes from a set of size $p$, and for each choice of $a$ there are $q$ choices for object $b$. Then the size of $S$ is $ p\times q:$}\\$$|S|= p \times q.$$
\end{frame}
\section{}
\begin{frame}
\frametitle{Four Basic Counting Principles}
{\Large \bf MULTIPLICATION PRINCIPLE :}
{\it Let $S$ be a set of ordered pair $ (a,b) $ of objects, where the first object $a$ comes from a set of size $p$, and for each choice of $a$ there are $q$ choices for object $b$. Then the size of $S$ is $ p\times q:$}\\$$|S|= p \times q.$$
The multiplication principle is actually a consequence of the addition principle. Let $a_1,a_2,\dots,a_p$ be the $p$ different choices for the object $a$. We partition $S$ into parts $S_1,S_2
,\dots,S_p$ where $S_i$ is the set of ordered pairs in $S$ with first object $a_i$, ($i$=1,2,\dots,$p$). The size of each $|S_i|$ is $q$; hence, by the addition principle,
\begin{center}
$|S|=|S_1|+|S_2|+\dots+|S_p|$\\
$=q+q+\dots+q$\\
$ =p\times q$
\end{center}
\end{frame}
\begin{frame}
\frametitle{Four Basic Counting Principles}
{\bf Example 1: }A student is to take two courses. The first meets at any one of 3 hours in the morning, and the second at any one of 4 hours in the afternoon. The number of schedules that are possible for the student is $ 3 \times 4 = 12 $.\\
\end{frame}
\begin{frame}
\frametitle{Four Basic Counting Principles }
{\bf Example 1: }A student is to take two courses. The first meets at any one of 3 hours in the morning, and the second at any one of 4 hours in the afternoon. The number of schedules that are possible for the student is $ 3 \times 4 = 12 $.\\
{\bf Example 2: }Chalk comes in three different length, 8 different colors, and 4 different diameters. How many different kind of chalk are there ?\\
{\bf Ans. }To determine a piece of chalk we carry out 3 different task : choose a length, choose a color, choose a diameter. By the multiplication principle, there are $ 3\times8\times4 = 96 $ different kinds of chalk.
\end{frame}
\section{}
\begin{frame}
\frametitle{Four Basic Counting Principles}
{\Large \bf SUBTRACTION PRINCIPLE : }
{\it Let $A$ be a set and let $U$ be a larger set containing $ A$. Let
$$\overline{A}= \{ x\in U: x \not\in A\} $$
be the complement of $ A$ in $U$. Then the number $|A|$ of objects in $A$ is given by the rule
$$|A|=|U|-|\overline{A}|.$$}
\end{frame}
\section{}
\begin{frame}
\frametitle{Four Basic Counting Principles}
{\Large \bf SUBTRACTION PRINCIPLE : }
{\it Let $A$ be a set and let $U$ be a larger set containing $ A$. Let
$$\overline{A}= \{ x\in U: x \not\in A\} $$
be the complement of $ A$ in $U$. Then the number $|A|$ of objects in $A$ is given by the rule
$$|A|=|U|-|\overline{A}|.$$}
In applying the subtraction principle, the set $U$ is usualy some natural set consisting of all the objects under discussion (the so-called universal set). Using the subtraction principle makes sense only if it is esier to count the number of objects in $U$ and $|\overline{A}|$ then to count the number of objects in $A$.
\end{frame}
\section{}
\begin{frame}
\frametitle{Four Basic Counting Principles}
{\bf Example 1: }Count the number of integers between 1 and 600, which are not divisible by 6.\\
{\bf Ans . } Here $U$ = the whole set = 600 \\
$A$ = \{$x:x$ is between 1 and 600 and not divisible by 6 \}
\begin{eqnarray*}
\overline{A} &=& \{x : x \not\in A \} \\
&=& \{ x : x/6 \}\\
&=&600/6\\
&=& 100\\
therefore,\\
|A| &=& |U|-\overline{|A|} \\
&=&600-100\\
&=&500\\
\end{eqnarray*}
\end{frame}
\section{}
\begin{frame}
\frametitle{Four Basic Counting Principles}
{\Large \bf DIVISON PRINCIPLE :}
{\it Let $S$ be a finite set that is partitioned into $k$ parts in such a way that each part contains the same number of objects. Then the number of parts in the partition is given by the rule \\
$$k = \frac{|S|}{ number \hspace{.2cm} of \hspace{.2cm}objects\hspace{.2cm} in\hspace{.2cm} a\hspace{.2cm} part }$$}
\end{frame}
\section{}
\begin{frame}
\frametitle{Four Basic Counting Principles}
{\Large \bf DIVISON PRINCIPLE :}
{\it Let $S$ be a finite set that is partitioned into $k$ parts in such a way that each part contains the same number of objects. Then the number of parts in the partition is given by the rule \\
$$k = \frac{|S|}{ number \hspace{.2cm} of \hspace{.2cm}objects\hspace{.2cm} in\hspace{.2cm} a\hspace{.2cm} part }$$}
Thus, we can determine the number of parts if we know the number of objects in $S$
and the common value of the number of objects in the parts.{\vspace{1cm}}
{\bf Example : } There are 740 pigeons in a collection of pigeonholes. If each pigeonhole contains 5 pigeons, the number of pigeonholes equals\\
$$740/5 =148$$
\end{frame}
\section{}
\begin{frame}
\frametitle{Permutations}
Let $r$ be a positive integer. By an $r$-$permutation$ of a set $S$ of $ n$ elements, we understand an ordered arrangement of $ r$ of the $n$ elements.
\end{frame}
\section{}
\begin{frame}
\frametitle{Permutations}
Let $r$ be a positive integer. By an $r$-$permutation$ of a set $S$ of $ n$ elements, we understand an ordered arrangement of $ r$ of the $n$ elements. If
$$S = \{ a , b ,c \} $$
then the 1-$permutation$ of $S$ are
$$a \hspace{2cm} b \hspace{2cm} c$$
\end{frame}
\section{}
\begin{frame}
\frametitle{Permutations}
Let $r$ be a positive integer. By an $r$-$permutation$ of a set $S$ of $ n$ elements, we understand an ordered arrangement of $ r$ of the $n$ elements. If
$$S = \{ a , b ,c \} $$
then the 1-$permutation$ of $S$ are
$$a \hspace{2cm} b \hspace{2cm} c$$
\\the six 2-$permutations$ of $S$ are
$$ab \hspace{1cm} ac \hspace{1cm} ba \hspace{1cm} bc \hspace{1cm}ca \hspace{1cm} cb $$
\end{frame}
\section{}
\begin{frame}
\frametitle{Permutations}
Let $r$ be a positive integer. By an $r$-$permutation$ of a set $S$ of $ n$ elements, we understand an ordered arrangement of $ r$ of the $n$ elements. If
$$S = \{ a , b ,c \} $$
then the 1-$permutation$ of $S$ are
$$a \hspace{2cm} b \hspace{2cm} c$$
\\the six 2-$permutations$ of $S$ are
$$ab \hspace{1cm} ac \hspace{1cm} ba \hspace{1cm} bc \hspace{1cm}ca \hspace{1cm} cb $$
the six 3-$permutations$ of $S$ are
\\
$$abc \hspace{1cm}acb \hspace{1cm} bac \hspace{1cm} bca \hspace{1cm} cab \hspace{1cm} cba$$
\end{frame}
\section{}
\begin{frame}
\frametitle{Permutations}
{\bf Theorem 1 : }For $n$ and $r$ positive integers with $ r \le n$.\\
$$P( n , r) = n \times (n-1) \times ......\times (n - r + 1)$$
\end{frame}
\section{}
\begin{frame}
\frametitle{Permutations}
{\bf Theorem 1 : }For $n$ and $r$ positive integers with $ r \le n$.\\
$$P( n , r) = n \times (n-1) \times ......\times (n - r + 1)$$
{\bf Results :} For a non-negative integer n, we define $n$! by
$$n!=n\times(n-1)\times\dots\times2\times1$$
with the convention that 0!=1. We may then write
$$P(n,r)=\frac{n!}{(n-r)!}$$
\end{frame}
\section{}
\begin{frame}
\frametitle{Permutations}
{\bf Theorem 1 : }For $n$ and $r$ positive integers with $ r \le n$.\\
$$P( n , r) = n \times (n-1) \times ......\times (n - r + 1)$$
{\bf Results :} For a non-negative integer n, we define $n$! by
$$n!=n\times(n-1)\times\dots\times2\times1$$
with the convention that 0!=1. We may then write
$$P(n,r)=\frac{n!}{(n-r)!}$$
{\bf Example 1 : } The number of 4-letter ``words" that can be formed by using each of letters $a$, $b$, $c$, $d$, $e$ at most once is $P(5,4)$, and this equals 5!/ (5-4)! = 120. The number of 5-letter words equals $P(5,5)$, which is also 120 .
\end{frame}
\section{}
\begin{frame}
\frametitle{Permutations}
{\bf Theorem 2: } The number of circular $r$-permutations of a set of n elements is given by
$$\frac{P(n,r)}{r} = \frac{n!}{r\times(n-r)!}$$
In particular, the number of circular permutations of $n$ elements is $(n-1)!$
\end{frame}
\section{}
\begin{frame}
\frametitle{Permutation}
{\bf Theorem 2: } The number of circular $r$-permutations of a set of n elements is given by
$$\frac{P(n,r)}{r} = \frac{n!}{r\times(n-r)!}$$
In particular, the number of circular permutations of $n$ elements is $(n-1)!$
{\bf Example :}
\end{frame}
\section{}
\begin{frame}
\frametitle{Permutations}
There are some results in permutation. they are {\vspace{.5cm}}
\begin{enumerate}
\item $P(n,n)=n!${\vspace{.3cm}}
\item $P(n,0)=\frac{n!}{(n-0)!}=1${\vspace{.3cm}}
\item $P(n,1)=1${\vspace{.3cm}}
\item $P(n,n-1)=n!$
\end{enumerate}
\end{frame}
\section{}
\begin{frame}
\frametitle{Permutation of Multiset}
If $S$ is a multiset, an $r$-$permutation$ of $S$ is an ordered arrangement of $r$ of the objects of $S$. If the total number of objects of $S$ is $n$ (counting repetitions), then an $n$-permutation of $S$ will also be called a $permutation$ of $S$.
\end{frame}
\section{}
\begin{frame}
\frametitle{Permutation of Multiset}
If $S$ is a multiset, an $r$-$permutation$ of $S$ is an ordered arrangement of $r$ of the objects of $S$. If the total number of objects of $S$ is $n$ (counting repetitions), then an $n$-permutation of $S$ will also be called a $permutation$ of $S$. For example, if $S$= $\{2.a, 1.b, 3.c\}$, then
$$acbc \hspace{1cm} cbcc$$
are 4-permutations of $S$, while
$$abccca$$
is a permutation of $S$.
\end{frame}
\section{}
\begin{frame}
\frametitle{Permutation of Multiset}
If $S$ is a multiset, an $r$-$permutation$ of $S$ is an ordered arrangement of $r$ of the objects of $S$. If the total number of objects of $S$ is $n$ (counting repetitions), then an $n$-permutation of $S$ will also be called a $permutation$ of $S$. For example, if $S$= $\{2.a, 1.b, 3.c\}$, then
$$acbc \hspace{1cm} cbcc$$
are 4-permutations of $S$, while
$$abccca$$
is a permutation of $S$.\\{\bf Remark :} The multiset $S$ has no 7-permutations since 7 is greater then 2+1+3=6, the number of objects of $S$.
\end{frame}
\section{}
\begin{frame}
\frametitle{Permutation of Multiset}
We count the number of $r$-permutations of a multiset $S${\vspace{1cm}}
\begin{enumerate}
\item each of whose repetition number is infinite.{\vspace{1cm}}
\end{enumerate}
\end{frame}
\section{}
\begin{frame}
\frametitle{Permutation Of Multiset}
We count the number of $r$-permutations of a multiset $S${\vspace{1cm}}
\begin{enumerate}
\item each of whose repetition number is infinite.{\vspace{1cm}}
\item each of whose repetition number is finite.
\end{enumerate}
\end{frame}
\section{}
\begin{frame}
\frametitle{Permutation of Multiset}
{\bf Theorem 1:} Let S be a multiset with objects of k different types, where each has an infinite repetition number. Then the number of $r$-permutations of $S$ is $k^r$.
\end{frame}
\section{}
\begin{frame}
\frametitle{Permutation of Multiset}
{\bf Theorem 1:} Let S be a multiset with objects of k different types, where each has an infinite repetition number. Then the number of $r$-permutations of $S$ is $k^r$.\\{\vspace{1cm}}
{\bf Example 1:} What is the number of ternary numerals with at most 4 digits.\\
{\bf Ans. } The answer to this question is the number of 4-permutations of the multiset $\{ \infty.0, \infty.1, \infty.2 \}$ or of the multiset $ \{ 4.0, 4.1, 4.2 \} $. By previous Theorem, this number equals $3^4=81$.
\end{frame}
\section{}
\begin{frame}
\frametitle{Permutation of Multiset}
{\bf Theorem 2: }Let $S$ be a multiset with objects of $k$ different types with finite repetition numbers $n_1,n_2,\dots,n_k$, respectively. Let the size of $S$ be $n=n_1+n_2+\dots+n_k$. Then the number of permutations of $S$ equals
$$ \frac{n!}{n_1!n_2!\dots n_k!}.$$
\end{frame}
\section{}
\begin{frame}
\frametitle{Permutation of Multiset}
{\bf Theorem 2: }Let $S$ be a multiset with objects of $k$ different types with finite repetition numbers $n_1,n_2,\dots,n_k$, respectively. Let the size of $S$ be $n=n_1+n_2+\dots+n_k$. Then the number of permutations of $S$ equals
$$ \frac{n!}{n_1!n_2!\dots n_k!}.$$
{\bf Example 2: } The number of permutation of the letters in the word MISSISSIPPI is
$$\frac{11!}{1!4!4!2!},$$
since this number equals the number of permutations of the multiset $\{ 1.M, 4.I, 4.S, 2.P \}$
\end{frame}
\section{}
\begin{frame}
\frametitle{Combinations}
Let $r$ be a non negative integer. By an $r$-$combination$ of a set $S$ of $n$ elements, we understand an unordered selection of $r$ of the $n$ objects of $S$.
\end{frame}
\section{}
\begin{frame}
\frametitle{Combinations}
Let $r$ be a non negative integer. By an $r$-$combination$ of a set $S$ of $n$ elements, we understand an unordered selection of $r$ of the $n$ objects of $S$. If $S= \{a,b,c,d\}$, then
$$ \{a,b,c\} , \{a,b,d\} ,\{a,c,d\}, \{b,c,d\}$$
are the four 3-$cobination$ of $S$.
We denote by $\binom{n}{r}$ the number of $r$-combinations of an $n$-element set.
\end{frame}
\section{}
\begin{frame}
\frametitle{Combinations}
Let $r$ be a non negative integer. By an $r$-$combination$ of a set $S$ of $n$ elements, we understand an unorderd selection of $r$ of the $n$ objects of $S$. If $S= \{a,b,c,d\}$, then
$$ \{a,b,c\} , \{a,b,d\} ,\{a,c,d\}, \{b,c,d\}$$
are the four 3-$cobination$ of $S$.
We denote by $\binom{n}{r}$ the number of $r$-combinations of an $n$-element set. Obviously,
\\ $$\binom{n}{r} = 0 \hspace{1cm} if \hspace{.5cm} r>n.$$
also, $$\binom{0}{r}=0 \hspace{1cm} if \hspace{.5cm} r>0.$$
\end{frame}
\section{}
\begin{frame}
\frametitle{Combinations}
the following additional facts are readily seen to be true for each non-negative integer n:{\vspace{.8cm}}
\begin{enumerate}
\item $\binom{n}{0} = 1${\vspace{.4cm}}
\item $\binom{n}{1} = n${\vspace{.4cm}}
\item $\binom{n}{n} = 1 ${\vspace{.4cm}}
\item $\binom{0}{0} = 1${\vspace{.4cm}}
\end{enumerate}
\end{frame}
\section{}
\begin{frame}
\frametitle{Combinations}
{\bf Theorem :} For $ 0 \le r \le n,$
$$P(n,r)=r!{\binom{n}{r}}.$$
Hence, $$\binom{n}{r} = \frac{n!}{r!(n-r)!}.$$
\pause
{\bf Example :} Twenty-five points are chosen in the plane so that no three of them are collinear. How many straight lines do they determine?\\
{\bf Ans.} Since no three of the points lie on a line, every pair of points determines a unique straight line. Thus, the number of straight lines determined equals the number of 2-combinations of a 25-element set, and this is given by
$$\binom{25}{2}=\frac{25!}{2!23!}=300.$$
\end{frame}
\section{}
\begin{frame}
\frametitle{Combinations of Multiset}
If $S$ is a multiset, then an $r$-$combination$ of $S$ is an unordered selection of $r$ of the objects of $S$. Thus, an $r$-combination of $S$ is itself a multiset, a submultiset of $S$.
\begin{enumerate}
\item If $S$ has $n$ objects, then there is is only one $n$-combination of $S$, namely, $S$ itself.
\item If $S$ contains objects of $k$ different types, then there are $k$ 1-combinations of $S$.
\end{enumerate}
\end{frame}
\section{}
\begin{frame}
\frametitle{Combinations of Multiset}
If $S$ is a multiset, then an $r$-$combination$ of $S$ is an unordered selection of $r$ of the objects of $S$. Thus, an $r$-combination of $S$ is itself a multiset, a submultiset of $S$.
\begin{enumerate}
\item If $S$ has $n$ objects, then there is is only one $n$-combination of $S$, namely, $S$ itself.
\item If $S$ contains objects of $k$ different types, then there are $k$ 1-combinations of $S$.
\end{enumerate}
{\bf Example :} If $S$=$\{2.a, 1.b, 3.c \}$, then the 3-combinations of $S$ are
$$\{2.a,1.b\},\hspace{1cm}\{2.a,1.c\},\hspace{1cm}\{1.a,1.b,1.c\},$$
$$\{1.a,2.c\},\hspace{1cm}\{1.b,2.c\},\hspace{1cm}\{3.c\}.$$
\end{frame}
\section{}
\begin{frame}
\frametitle{Combinations of Multiset}
{\bf Theorem :} Let $S$ be a multiset with objects of $k$ types, each with an infinite repetition number. Then the number of $r$-combinations of $S$ equals
$$\binom{r+k-1}{r}=\binom{r+k-1}{k-1}$$
\end{frame}
\section{}
\begin{frame}
\frametitle{Combinations of Multiset}
{\bf Theorem :} Let $S$ be a multiset with objects of $k$ types, each with an infinite repetition number. Then the number of $r$-combinations of $S$ equals
$$\binom{r+k-1}{r}=\binom{r+k-1}{k-1}$$
{\bf Example :} A bakery boasts 8 varieties of doughnuts. If a box of doughnuts contain 1 dozen, how many different options are there for a box of doughnuts? \\
{\bf Ans. } It is assumed that the bakery has on hand a large number (at least 12) of each variety. This is a combination problem, since we assume the of order of the doughnuts in a box is a box is irrelevant for the purchaser's purpose. The number of different options for boxes equals the number of 12-combinations of a multiset with objects of 8 types, each having an infinite repetition number. This number equals \\
$$\binom{12+8-1}{12}=\binom{19}{12}$$
\end{frame}
\section{}
\begin{frame}
\frametitle{Conclusion}
We have discuss these following topics :{\vspace{.2cm}}
\begin{enumerate}
\item Using the Counting Principles{\vspace{.2cm}}
\item Finding the simple way of permutation{\vspace{.2cm}}
\item Solving the Problem of Permutation{\vspace{.2cm}}
\item Finding the simple way of permutation of Multiset{\vspace{.2cm}}
\item Finding the simple way of Combinations{\vspace{.2cm}}
\item Solving the Problem of Combinations{\vspace{.2cm}}
\end{enumerate}
\end{frame}
\section{References}
\begin{frame}
\frametitle{References}
\begin{thebibliography}{5}
\bibitem{beamer} \emph{ Richard A. Brualdi: $Introductory$ $Combinatorics$, Fourth edition. Madison: University of Wisconsin, 2008.}{\vspace{.6cm}}
\bibitem{beamer2} \emph{C.L.Liu: Elements of Discrete Mathemathics, Second Edition. Urbana-Champaign: University of Illinois, 2000 }{\vspace{.6cm}}
\bibitem{beamer2} \emph{Ralph P.Grimaldi: Discrete and Combinatorial Mathematics, Fifth edition.USA: Rose–Hulman Institute of Technology,2003}{\vspace{1cm}}
\end{thebibliography}
\end{frame}
\section{}
\begin{frame}
\frametitle{ Counting principle, Permutation and Combination}
%\includegraphics[width=\textwidth]{22.png}
%\includegraphics[width=\textwidth]{download.jpg}
\includegraphics[width=\textwidth]{Thank-You-Card.jpg}
\end{frame}
\end{document}