Discrete Mathematics with Graph Theory (3rd Edition) 214

Discrete Mathematics with Graph Theory (3rd Edition) 214 -...

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212 Solutions to Review Exercises 2. As a group, the girls can be arranged in 6! ways. Then, together with the boys, they form 8 "units" to be arranged in a row. Since the girls cannot appear first, there are seven choices for the first position in the row. Then there are six choices for the last position (again, not the girls). There are 6! ways to fill out the remaining positions, so the answer is 7(6)(6!)(6!) = 21,772,800. 3. The answer is zero! In a circle, if the six girls are together, then the boys must also be together. 4. Let A be the set of permutations containing aeb and B the set of permutations containing bef. We want IA UBI. By the Inclusion-Exclusion Principle, this is IAI + IBI-IA n BI. Now IAI = 4! (aeb is a single unit in a string of four "units"), IBI = 4! and IA n BI = 2! (if both aeb and bef appear, then aebef must appear). Thus IA U BI = IAI + IBI- IA n BI = 24 + 24 - 2 = 46. 5. (a) Let
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.

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