Discrete Mathematics with Graph Theory (3rd Edition) 186

# Discrete Mathematics with Graph Theory (3rd Edition) 186 -...

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184 Solutions to Exercises 11. Let A be the set of permutations which contain the pattern bge: IAI = 5! (thinking of bge as a single symbol). Similarly, if B is the set of permutations which contain eaj, IBI = 5!. For a permutation to contain both and eaj it must contain bgeaj, so IA n BI = 3!. The number containing either or is IA U = + IBI - IA n = 5! + 5! - 3! = 234. Hence, the answer is 7! - 234 = 5040 - 234 = 4806. 12. [BB] Let N be the set of lines in which the Noseworthys are beside each other and A the set of lines in which the Abbotts are beside each other. (a) For each of the two ways in which the Noseworthys can stand beside each other, there are 3! possible lines-the Noseworthys as one unit and the two Abbotts. Thus, INI = 2 x 3! = 12. (b) The number of possible lines altogether is 4! = 24. Thus, INcl = 24 - 12 = 12. (c) There are 2! orders for the couples and, in each case, two arrangements for the Noseworthys and two for the Abbotts: IN n AI = 2 x 2 x 2 = 8. (d) The Noseworthys must be in positions 2 and 3 in either order; the Abbotts must be in positions 1 and 4 in either order. There are four lines.
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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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