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By Ainouche A., Schiermeyer I.

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Chapter 1. Basic Methods 44 each team of the other conference, one has to be at home, one away. Finally, the Magic can choose five of the remaining opponents whom they will play twice at home, once away, and then they will play the remaining four opponents twice away, and once at home. How many different schedules can the Orlando Magic have? 21. How many ways are there to choose subsets Sand T of [n] if there are no conditions whatsoever imposed on these subsets? 22. (a) How many ways are there to choose subsets Sand T of [n] so that S contains T?

Continuing the line of thought of the solution of the previous exercise, we can then place our k rooks into the chosen columns in (n)k ways. Therefore, the total number of possibilities is 5. /2. 9. Solutions to Exercises that is, those in which Bob finishes ahead of Amy. Then there is a bijection f : S ----; T, namely the function that simply switches Amy and Bob. 6. As the order of the numbers played does not matter, this problem simply asks for the number of all six-element subsets of [49]. 4, this is (~) = 13983816.

We prove the theorem by showing that its opposite is impossible. That is, we assume that the opposite of our statement is true (that is, in this case, we assume that there is no index i so that IAil > r holds), and we derive a contradiction from this assumption. This procedure is called an indirect proof or proof by contradiction. Proof: Let us assume that the statement we want to prove is false. Then IAil ::; r holds for each i. Therefore, 36 Chapter 1. Basic Methods which contradicts our original assumption that IAI u A2 u· ..

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0-Dual Closures for Several Classes of Graphs by Ainouche A., Schiermeyer I.


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