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Various topics that we have studied

January 21, 2015

Keep this le for future reference. I will update it periodicaly.

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1 Counting

Before we discuss the ne art of counting, we need to discuss both some terminology andsome more basic concepts. For the purpose of these notes the following statements are true.

The words distinct and di ff erent are the synonyms. The words outcomes and possibilities are synonyms.

The words sequence and strings are synonyms. A sequence is dened as an ordered list of elements, which need not to be distinct. For example, R,U,S,T,E,A,L,P,H,O,L,O,W,I,N,G,M,Eis a sequence of letters.

A set is commonly dened as a collection of distinct elements (for example A = {C,U,L}and B = {Y,U,P } are sets), while objects that contain multiple copies are called mul-tisets.

The union of two sets (AUB) is the set of elements in either or both sets. For exampleAUB = {C,U,L}U {Y,U,P } = {C,U,L,Y,P }. So, the union of two sets is the smallestset which contains both sets as subsets.

The intersection of two sets is the set of elements the two have in common. In ourexample (AintB ) = {C,U,L}int {Y,U,P } = {U }. If there is no element that is commonto both sets, we say that the sets are disjoint and their intersection is the emptyset .

| A| denotes the number of elements in a set (not the absolute value in this case sincesometimes we want to make things a little confusing).

A partition of a set is a collection of disjoint subsets whose union equals the entire set.

Got it?! You imagine how easy math is since we call these concepts basics. But, hey, donot despair ... there is light at the end of the tunnel. Who knows, maybe using these abstractconcepts can help us become better people in life (whatever better means).

1.1 Basics of Counting

Most people think counting is easy. And you cant blame them since this is sometimes true.Consider counting the number of elements in a sequence of integers say from 0 to 10. It easyto verify that this sequence has 11 elements. More generally, if we have a sequence of integerssay from a to b (with the property than a < b) we can nd that, including a and b, there are

b a + 1 elements (to see this just consider the previous example: b = 10 and a = 0 so thenumber of elements is 10 0 + 1 = 11).

Although we didmt explicitly talked about them, there are two basic principles in countingthat you should be aware of (the formal denitions are more complicated but the basic ideasbehind them are easier to understand).

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1. Sum rule : suppose a certain task (or operation) can be done by two di ff erent procedures.One procedure has n possible outcomes (to be more formal you can say a set of outcomes)while the other has m outcomes (or a set of m outcomes). Moreover assume that theoutcomes of the two tasks are totally di ff erent - we say that the set of outcomes n and

the set of outcomes m are disjoint (we can also use an earlier basic concept: disjoint setsare sets whose ). Then, the number of possible outcomes is n+m.

This rule is a lot simpler than the fancy statement we used to describe it. Lets saymommy has decided to go shopping today :-( and she wants to take her favoriteson with her. However, she decided to shop only at one store :-), either in the northpart or the south part of town. If she decides to go to the north part she can stopeither at a clothing store, a jewelry store of a furniture store. If she goes southof town she can choose between a shoe store or arts and crafts store. How manypossible shops can mommy end shopping at today? The answer is straight forward3+2=5. Therefore there are 5 possible outcomes.

2. Product rule : When there are m ways to do one thing, and n ways to do another, thenthere are mn ways of doing both. This can be extended: if the are n1 ways to do onething, n2 to do another and n3 to do another, then there are n1 x n2 x n3 ways of doingall three.

Example: you have 5 shirts and 5 pants.That means 55=25 di ff erent outts. If wealso want to consider the 5 pairs of socks and 5 pairs of snickers that you have ...you have 5x5x5x5=625 di ff erent outts. Look in your closet ... how many di ff erentoutts can you wear? So stop wearing the same combination every day!

Unfortunately, sometimes counting gets more complicated. We learned that whenever we wantto count things can get tricky. We need to think very clearly about what exactly we want tocount. Do we want to allow for repetition? Do we think the order we count matters? Theseare all great questions.

As we discussed, the answer is quite di ff erent depending on the answer to these questions.Do we allow for repetition? If yes. then things get more complicated since we need to dealwith more outcomes. Do we care about he order we choose between possible outcomes? If wedo care, we will have to deal with more outcomes than if we dont. That is why we talkedabout permutations and combinations. Moreover, that is why we considered permutationswith and without repetitions.

Before we formally dene these concepts, I would like to give you two examples (or waysto think about these concepts). Suppose you are choosing a password for you email account.Furthermore, suppose you can choose only from 1, 2 and 3 to come up with a 3-digit password.Suppose you chose 321. Then, the order is very important since this is the only sequence thatyou can use to log in. This an example of PERMUTATIONS - ORDER DOES MATTER!

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Next, consider the following example: you want to make a salad using lettuce, tomatoesand cucumbers. Is the order you throw them in the bowl important? Most likely not. Youare going to eat the same salad no matter what order you choose.

To introduce the concepts of permutations and combinations I am going to work with a

particular example. Suppose we have 3 numbers 1, 2, 3 and we need to choose all or some of them to form 3 or 2-digit numbers.

1.2 Permutations

When computing permutations, as I mentioned, we care about order. However, we can alsoconsider repetitions.

Lets consider our example. Suppose we have three numbers 1,2, 3 and we need to choose2 of them to form a 2-digit number. Furthermore, suppose rst we are allowed to use thesame number when coming up with distinct 2-digits numbers (translation: order does matterand we allow for repetition). So lets see how many possibilities we have: (1,1), (1,2), (1,3),(2,1), (2,2),(2,3),(3,1),(3,2),(3,3). So 9 of them. We can generalize this result. Let n=numberof objects we have (in our example the three numbers: 1,2 and 3) and r=number of objectschosen from all the objects we have (in our case two). Then, the number of permutations withrepetition is nr (in our example 3 2 =9). So, if we want to nd the number of permutationswith repetition that we can form by choosing r possible objects from n objects we canuse the following formula: nr .

On the other hand, if we do no allow for repetition we have a very di ff erent formula. Letstake our example we have three numbers 1,2, 3 and we need to choose 2 of them to form a2-digit number but we are not allow to repeat any digit. Now for he rst digit of the 2-digitnumber we have 3 choices: 1,2,3. However, given that we chose the rst digit, for the seconddigit we only have 2 choices. So we have 3x2 = 6choices.We actually have a general formulafor this types of permutations (without repetition): nP r = n!/ (n r)!.

1.3 Combinations

Algebra

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