hw6math109

# hw6math109 - MATH 109 Assignment 6 Due#12.4 Solution Let X...

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Unformatted text preview: MATH 109 Assignment 6 Due: 5/16/11 #12.4 Solution. Let X, Y, Z be the possible dishes that each of the three diners can choose respectively. Since the menu consists of five items we have | X | = | Y | = | Z | = 5. The set X × Y × Z represents the set of choices possible if we record who selected which dish. Hence, by the multiplication principle (Theorem 10.2.3), there are | X × Y × Z | = | X || Y || Z | = 5 3 choices. If we do not care who orders which dish, then there are (i) 5 ways for the diners to choose one dish from the menu, (ii) ( 5 2 ) · 2 = 20 ways for the diners to choose two dishes from the menu, since one dish is chosen once and the other dish is chosen twice, (iii) ( 5 3 ) = 10 ways for the diners to choose three dishes from the menu. Hence, in total, there are 35 choices if we do not care who orders which dish. #14.1 Proof. Since A is finite, we have A \ B is too, so write A \ B = { x 1 . . . , x n } . Observe that A ∪ B = ( A \ B ) ∪ B and that A \ B and B are disjoint. Asare disjoint....
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## This note was uploaded on 08/02/2011 for the course MATH 109 taught by Professor Knutson during the Spring '06 term at UCSD.

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hw6math109 - MATH 109 Assignment 6 Due#12.4 Solution Let X...

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