This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: a Øh4Ô Ö iÒg6ÓÑ eÛ Ó Ök# Ó ÐÙ Ø iÓÒ ×bÝ iÑC a ÐÐahaÒ § 4.3 Problem 18 ( a ): There are 2 8 = 256 possible outcomes. ( b ): There are parenleftbigg 8 3 parenrightbigg = 8! 3!5! = 56 outcomes with three heads. ( c ): The number with fewer than three heads is parenleftbigg 8 parenrightbigg + parenleftbigg 8 1 parenrightbigg + parenleftbigg 8 2 parenrightbigg = 1 + 8 + 28 = 37 , so there are 219 with at least three heads. ( d ): To have the same number of heads as tails there must be four of each. Thus the answer is parenleftbigg 8 4 parenrightbigg = 70 . § 4.3 Problem 23: First we position the eight men. Then there are nine spaces between and at the ends. Each of these spaces can hold at most one woman, so we just pick which five of these nine have women in them. There are ( 9 5 ) ways to do this. Then we consider that there are 8! ways of ordering the men and 5! ways of ordering the women. Thus the number of such orderings is parenleftbigg 9 5 parenrightbigg · 8! · 5! = 609 638 400 . § 4.3 Problem 25: In order to match the solutions in the back of the book we must assume that the drawings are made without replacement. That is, once ticket number 37 is drawn it is not put back in the hat, so nobody can win more than one prize. ( a ): There are 100 choices for the first prize, 99 for the second, etc. , so the answer is 100! (100 4)! = 100 · 99 · 98 · 97 = 94 109 400 ....
View
Full
Document
This note was uploaded on 07/12/2008 for the course MAT 243 taught by Professor Callahan during the Spring '06 term at ASU.
 Spring '06
 CALLAHAN

Click to edit the document details