Combinatorial Analysis - download pdf or read online

By R. Bellman, M. Hall (ed.)

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Thus the number of different shopping trips is given by |A1 ||A2 ||A3 ||A4 ||A5 | = 4(3)(2)(3)(3) = 216 by the Multiplication Principle. ✷ Suppose that we want to make a string of n colored beads. Each bead may be one of m colors and we have unlimited beads of each color. As usual, we want to determine the number of visually distinct strings. Usually, we would only be interested in visually distinct strings, in other words, those that cannot be obtained from another by flipping the string. So, if we are using the colors 1, 2, and 3, then 1223 and 3221 would be considered the same string.

8 until Chap. 7. , An . , An are mutually disjoint sets, then: n | ∪ni=1 Ai | = |Ai |. i=1 Proof We proceed by induction on n. If n = 1, then the claim is obvious. ,An satisfy: n | ∪ni=1 Ai | = |Ai |. ,An , and An+1 be mutually disjoint sets. Thus, ∪ni=1 Ai and An+1 are disjoint sets. 6, we have that n | ∪n+1 i=1 Ai | = | ∪i=1 Ai | + |An+1 |. 30 2 Basic Counting Applying the inductive hypothesis yields n | ∪n+1 i=1 Ai | = | ∪i=1 Ai | + |An+1 | n n+1 |Ai | + |An+1 | = = i=1 |Ai |. i=1 Alternatively, if x ∈ Ai , then x ∈ / Aj for j = i.

Find the number of valid passwords. Solution As a first attempt at a solution, we might try to put the letters in order from left to right. It is easy to see we have 26 possibilities for the first letter, 25 for the second (anything but the first letter used), and so on. However, placing the last letter is more difficult, as we do not know how many vowels have been used. To find a solution, we begin with the most restrictive selection we have to make. Namely, we begin by selecting the last letter.

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