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Published byθάλαΟΟΞ± ΞΞΈΞ±Ξ½Ξ±ΟΞΉΞ¬Ξ΄Ξ·Ο Modified over 5 years ago
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Discrete Math for CS CMPSC 360 LECTURE 27 Last time: Counting. Today:
Combinatorial proofs. CMPSC 360 5/11/2019
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1st rule of counting If an object can be formed by a succession of π choices, where there are π 1 ways of making the first choice, and for every way of making the 1st choice, there are π 2 ways of making the 2nd choice, and for every way of making the 1st and 2nd choices there are π 3 ways of making the 3rd choice, and so on up to the π π -th choice, then the number of distinct objects that can be made in this way is π 1 β
π 2 β
π 3 β
β¦β
π π . 5/11/2019
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2nd rule of counting Suppose an object can be formed by a succession of choices, and the order in which the choices are made does not matter. Let A be the set of possible objects if the order of choices matters. Let B the set of possible objects if it does not. Suppose we can map all objects from A to objects in B is that exactly π‘ elements of A are mapped to every element of B. Then |π΅|= π΄ π‘ . 5/11/2019
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Typical counting problems
Order π elements: π! ways. Choose π out of π without replacement, ordered: π! πβπ ! ways. Choose π out of π without replacement, unordered: π π = π! π! πβπ ! ways. Choose π out of π with replacement, ordered: π π ways. Choose π out of π with replacement, unordered: ??? 5/11/2019
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I-clicker question (frequency: BC)
How many undirected graphs with vertex sets 1,2,β¦,π are there? π 2 2 π 2 π πβ1 /2 2 π πβ1 2 π 2 5/11/2019
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I-clicker question (frequency: BC)
How many directed graphs with vertex sets 1,2,β¦,π are there? π 2 2 π 2 π πβ1 /2 2 π πβ1 2 π 2 5/11/2019
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Typical counting problems
Order π elements: π! ways. Choose π out of π without replacement, ordered: π! πβπ ! ways. Choose π out of π without replacement, unordered: π π = π! π! πβπ ! ways. Choose π out of π with replacement, ordered: π π ways. Choose π out of π with replacement, unordered: π+πβ1 π ways. 5/11/2019
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A combinatorial proof To prove an equality of the form LHS=RHS:
Find a counting question you can answer in two ways. Argue that the first way gives LHS. Argue that the second way gives RHS. Conclude that LHS=RHS. It is like playing Jeopardy!. Ex.1. π π+1 = πβ1 π + πβ2 π +β¦+ π π Ex.2. π π = πβ1 πβ1 + πβ1 π 5/11/2019
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