4 1 Basic principles Multiplication principle If an
4. 1 Basic principles Multiplication principle If an activity can be performed in k successive steps, n n Step 1 can be done in n 1 ways Step 2 can be done in n 2 ways … Step k can be done in nk ways Then: the number of different ways that the activity can be performed is the product n 1 n 2…nk
Addition principle Let X 1, X 2, …, Xk be a collection of k pairwise disjoint sets, each of which has nj elements, 1 < j < k, then the union of those sets k X= Xj j =1 has n 1 + n 2 + … + n k elements
4. 2 Permutations and combinations A permutation of n distinct elements x 1, x 2, …, xn is an ordering of the n elements. There are n! permutations of n elements. Example: there are 3! = 6 permutations of three elements a, b, c: abc bac cab acb bca cba
r-permutations An r-permutation of n distinct elements is an ordering of an r-element subset of the n elements x 1, x 2, …, xn Theorem 4. 2. 10: For r < n the number of r-permutations of a set with n distinct objects is P(n, r) = n(n-1)(n-2)…(n-r+1)
Combinations Let X = {x 1, x 2, …, xn} be a set containing n distinct elements q An r-combination of X is an unordered selection of r elements of X, for r < n q The number of r-combinations of X is the binomial coefficient C(n, r) = n! / r!(n-r)! = P(n, r)/ r!
4. 7 Binomial coefficients q Theorem 4. 7. 1: Binomial theorem. For any real numbers a, b, and nonnegative integer n: (a+b)n = C(n, 0)anb 0 + C(n, 1)an-1 b 1 + … + C(n, n-1)a 1 bn-1 + C(n, n)a 0 bn
Generalized permutations Theorem 4. 6. 2: Suppose that a sequence of n items has nj identical objects of type j, for 1< j < k. Then the number of orderings of S is ____n!____ n 1!n 2!. . . nk!
Generalized combinations Theorem 4. 6. 5: Suppose that X is a set containing t distinct elements. Then the number of unordered, k-element selections from X, repetitions allowed, is C(k + t -1, t -1) = C(k + t -1, k)
Pascal’s Triangle 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 84 36 9 1 …
4. 8 The pigeonhole principle p First form: If k < n and n pigeons fly into k pigeonholes, some pigeonhole contains at least two pigeons.
Second form of the pigeonhole principle p If X and Y are finite sets with |X| > |Y| and f : X Y is a function, then f(x 1) = f(x 2) for some x 1, x 2 X, x 1 x 2.
Third form of the pigeonhole principle p If X and Y are finite sets with |X| = n, |Y| = m and k = n/m , then there at least k values a 1, a 2, …, ak X such that f(a 1) = f(a 2) = … f(ak). Example: n = 5, m = 3 k = n/m = 5/3 = 2. n
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