diff --git a/Intermediate/Combinatorics/main.tex b/Intermediate/Combinatorics/main.tex index d024d71..c205967 100755 --- a/Intermediate/Combinatorics/main.tex +++ b/Intermediate/Combinatorics/main.tex @@ -14,7 +14,7 @@ } } -\uptitlel{Intermediate 2} +\uptitlel{Advanced 1} \uptitler{ORMC Summer Sessions} \title{Combinatorics} \subtitle{Prepared by Mark on \today} @@ -25,13 +25,13 @@ \section{Getting started} - An ordered arrangement of objects is called a \textit{permutation}. An unordered selection\footnotemark of objects is called a \textit{combination}. All the following problems involve permutations.\\ + An ordered arrangement of objects is called a \textit{permutation}. An unordered selection\footnotemark{} of objects is called a \textit{combination}. All the following problems involve permutations. - \footnotetext{A ``combination lock'' cares about the order of its digits, so its name is inaccurate. Such an object is actually a \textit{permutation} lock!} + \footnotetext{A \say{combination lock} cares about the order of its digits, so its name is inaccurate. Such an object is actually a \textit{permutation} lock!} \problem{} - How many different ways are there to rearrange the letters ABCDE? \\ + How many different ways are there to rearrange the letters ABCDE? \begin{solution} $5 \times 4 \times 3 \times 2 \times 1 = 120$ @@ -40,14 +40,14 @@ \vfill \problem{} - How many different ways are there to arrange the letters ABCDEFG...XYZ? \\ - The answer is a \textit{very} big number. You should not fully resolve your answer. \\ + How many different ways are there to arrange the letters ABCDEFG...XYZ? \par + The answer is a \textit{very} big number. You should not fully resolve your answer. \par \hint{When you see a problem that's as big as this one, it's often wise to try and understand a simpler case first. Look at \ref{ABCDE} again, and try to create a general strategy.} \begin{instructornote} - A hint for students that are stuck: \\ - In \ref{ABCDE}, start with five blank spaces. How many choices are there for A's position? \\ + A hint for students that are stuck: \par + In \ref{ABCDE}, start with five blank spaces. How many choices are there for A's position? \par Once A is placed, how many are left for B? \end{instructornote} @@ -55,7 +55,7 @@ \pagebreak \definition{} - The \textit{factorial} of a positive integer $x$ is $x \times (x-1) \times ... \times 1$. We denote this $x!$. \\ + The \textit{factorial} of a positive integer $x$ is $x \times (x-1) \times ... \times 1$. We denote this $x!$. \par For example, $8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320$. \problem{} @@ -67,8 +67,8 @@ \vfill \problem{} - How many ways are there to choose three student council officers from a class of 20 students? \\ - How many ways are there to choose a president, a vice-president, and a treasurer from the same class? \\ + How many ways are there to choose three student council officers from a class of 20 students? \par + How many ways are there to choose a president, a vice-president, and a treasurer from the same class? \par \hint{You answers should be different. In which case does order matter?} \begin{instructornote} @@ -79,10 +79,10 @@ \pagebreak \problem{} - Say you have 4 red balls and 3 green balls. How many different ways can you arrange them on the table in font of you? \\ + Say you have 4 red balls and 3 green balls. How many different ways can you arrange them on the table in font of you? \par \begin{solution} - Consider the sequence RRRRGGG. There are $4!$ ways to rearrange the red balls, and $3!$ ways to rearrange the green balls. This is true for any sequence. \\ + Consider the sequence RRRRGGG. There are $4!$ ways to rearrange the red balls, and $3!$ ways to rearrange the green balls. This is true for any sequence. So, our solution is $\frac{7!}{3!4!}$. \end{solution} @@ -105,7 +105,7 @@ \section{Permutations} - It would be convenient to have a general tool for counting permutations. Let us try to create one. \\ + It would be convenient to have a general tool for counting permutations. Let us try to create one. \par (Remember, permutations are \textit{ordered} arrangements of objects.) First, let's create a function $\npk{n}{k}$, which tells us how many $k$-object permutations we can choose from a group of $n$ objects. @@ -116,7 +116,7 @@ \vfill - ``Choosing $k$ items from $n$'' is a lot like splitting our $n$ objects into two groups: those we choose, and those we don't. \\ + ``Choosing $k$ items from $n$'' is a lot like splitting our $n$ objects into two groups: those we choose, and those we don't. \begin{center} \begin{tikzpicture} @@ -146,7 +146,7 @@ If we rearrange these, we get different permutations. How can we count them? \problem{} - Using the above diagram, create a formula for $\npk{n}{k}$. \\ + Using the above diagram, create a formula for $\npk{n}{k}$. \par \hint{We're counting \textit{permutations}, so the order of items in the first group matters.} \begin{solution} @@ -159,9 +159,9 @@ \vfill \pagebreak - \section{Combinations} + \section{Combinations}ow m - Now, let's count \textit{combinations}. \\ + Now, let's count \textit{combinations}. \par Here, we care about \textit{which} items we choose, but not \textit{how} we choose them. We'll make a function $\nck{n}{k}$, or \textit{``n choose k''}. This will tell us how many different ways we can choose $k$ items from a set of $n$. \problem{} @@ -172,29 +172,43 @@ Usually, $\nck{n}{k}$ is written as $\binom{n}{k}$. This is also called the \textit{binomial coefficient}. + + \problem{} + Say we have a bag with the following contents: + \begin{itemize} + \item 8 identical blue balls + \item 3 identical green balls + \item 6 identical red balls + \item 4 identical pink balls + \end{itemize} + How many different ways are there to select 5 balls from this bag? + + + \problem{} + Now, derive the \textit{multinomial coefficient} $\binom{n}{k_1,k_2,...,k_m}$. \par + \vspace{1mm} + The multinomial coefficient tells us how many distinct ways we can choose $n$ objects from a set which has $m$ classes, and where each class $i$ contains $k_i$ identical objects. \par + \hint{In \ref{manyballs}, $n = 5$ and $(k_1, k_2, k_3, k_4) = (8, 3, 6, 4)$. \\ + So, the solution to \ref{manyballs} should be given by the multinomial coefficient $\binom{5}{8,3,6,4}$.} + + + + + + \vfill + \pagebreak + \section{Applications} - \problem{} - Use the meaning of \nck{n}{k} to explain why $\nck{n}{k} = \nck{n}{n - k}$ - - \vfill - - \problem{} - Use the formula of $\nck{n}{k}$ to explain why $\nck{n}{k} = \nck{n}{n - k}$ - - \vfill - \problem{} How many ways can a class of 27 people be seated in 30 seats? \vfill - \pagebreak - \problem{} - The following is the map of a city. Each line is a one-way road, you can only drive up or right. \\ - How many different paths can you take from A to B? \\ - How many of them go through the center point? \\ + The following is the map of a city. Each line is a one-way road, you can only drive up or right. \par + How many different paths can you take from A to B? \par + How many of them go through the center point? \begin{tikzpicture} \draw [step=0.5,gray] (0,0) grid (7*0.5,4*0.5); @@ -220,12 +234,12 @@ How many ways can we split the number 2016 into a sum of positive integers? \vfill + \pagebreak \problem{} A staircase must be built up a wall. It will start 4.5 meters away from the wall, which is 1.5 meters tall. The height of each step is exactly 30 centimeters. The width of each step must be an integer multiple of 50 centimeters. In how many ways can the staircase be constructed? \vfill - \pagebreak \section{Bonus problems} @@ -239,10 +253,4 @@ \vfill - \problem{} - A stressed-out student consumes at least one espresso every day of a particular year, drinking $500$ overall. Prove that on some consecutive sequence of whole days the student drinks exactly $100$ espressos. - \note[Warning]{This problem is significantly harder than anything else in the handout.} - - \vfill - \end{document}