# lect4_2 - Error Correcting Codes Combinatorics Algorithms...

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Error Correcting Codes: Combinatorics, Algorithms and Applications (Fall 2007) Lecture 4: Probability and Discrete Random Variables Wednesday, January 21, 2009 Lecturer: Atri Rudra Scribe: Anonymous 1 Counting and Probability This lecture reviews elementary combinatorics and probability theory. We begin by first reviewing elementary results in counting theory, including standard formulas for counting permutations and combinations. Then, the axioms of probability and basic facts concerning probability distributions are presented. 2 Counting Counting theory tries to answer the question ”How many?” or ”How many orderings of n distinct elements are there?” In this section, we review the elements of counting theory. A set of items that we wish to count can sometimes be expressed as a union of disjoint sets or as a Cartesian product of sets. The rule of sum says that the number of ways to choose an element from one of two disjoint sets is the sum of the cardinalities of the sets. That is, if A and B are two finite sets with no members in common, then | A B | = | A | + | B | . The rule of product says that the number of ways to choose an ordered pair is the number of ways to choose the first element times the number of ways to choose the second element. That is, if A and B are two finite sets, then | A × B | = | A |·| B | . A string over a finite set S is a sequence of elements of S . We sometimes call a string of length k a k-string . A substring s of a string s is an ordered sequence of consecutive elements of s . A k-substring of a string is a substring of length k . For example, 010 is a 3 -substring of 01101001 (the 3 -substring that begins in position 4 ), but 111 is not a substring of 01101001 . A k -string over a set S can be viewed as an element of the Cartesian product S k of k -tuples; thus, there are | S | k strings of length k . For example, the number of binary k -strings is 2 k . Intuitively, to construct a k -string over an n -set, we have n ways to pick the first element; for each of these choices, we have n ways to pick the second element; and so forth k times. This construction leads to the k -fold product n · n · · · n = n k as the number of k -strings. A permutation of a finite set S is an ordered sequence of all the elements of S , with each element appearing exactly once. For example, if S = { a, b, c } , there are 6 permutations of S : abc, acb, bac, bca, cab, cba. There are n ! permutations of a set of n elements, since the first element of the sequence can be chosen in n ways, the second in n - 1 ways, the third in n - 2 ways, and so on. 1

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A k-permutation of S is an ordered sequence of k elements of S , with no element appearing more than once in the sequence. Thus, an ordinary permutation is just an n -permutation of an n -set. The twelve 2 -permutations of the set { a, b, c, d } are ab, ac, ad, ba, bc, bd, ca, cb, cd, da, db, dc, where we have used the shorthand of denoting the 2 -set { a, b } by ab , and on on. The number of k -permutations of an n -set is n ( n - 1)( n - 2) · · · ( n - k + 1) = n !
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