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Unformatted text preview: CSE 21: Homework Solutions 2 October 5, 2009 Problem 1 Teams A and B play in baseballs world series. Here, the team that first wins four games wins the series. (a) What is the number of ways the series can occur? (b) What is the number of ways the series can occur, given that A wins the first game, and that the team which wins the second game also wins the fourth game? Solution: (a) Suppose there is in total 4 n 7 games in the series. Since the winner of the series must win the last game, the number of ways the series which has exact n games can occur is ( n 1 3 )( 2 1 ) . Thus, there is in total 7 X n =4 n 1 3 2 1 ways the series can occur. Another solution: Suppose w.l.o.g. that A wins. there are ( 7 4 ) sequences with 3 B s and 4 A s. This is in bijection with the tournament outcomes by truncating each individual sequence after the last A . Then, multiply this by 2 to cover the case that B wins, gives 2 ( 7 4 ) total ways. Note that this is equivalent to the above answer by using the identity n j = k ( k j ) = ( n +1 k +1 ) . (b) By using the Decision Tree method, we have the following 15 ways: AAAA, AABAA, AABABA, AABABBA, AABABBB, ABABAA, ABABABA, ABABABB, ABABBAA, ABABBAB, ABABBB, ABB BAAA, ABBBAAB, ABBBAB, ABBBB. Another solution: Look at string as AxyxS , where x,y are A or B and S is a string of A and B ....
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This note was uploaded on 02/02/2012 for the course CSE 21 taught by Professor Graham during the Fall '07 term at UCSD.
 Fall '07
 Graham

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