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Unformatted text preview: 1. In the expansion of ( x + y + 2 z ) 6 what is the coefficient in front of the term x 2 yz 3 . 2 3 6! 2!1!3! = 480 2. Starting from a point ( a,b ) you may move to the point ( a + 1 ,b + 1) or ( a +1 ,b 1). In other words as you move one to the right you must move either up one or down one as well. Starting from the point (0 , 0) how many different paths are there to the point (10 , 6)? This is the same as moving up 8 times and down 2 as we move to the right 10 times. Hence the number of different paths is ( 10 8 ) = 45. Starting from the point (10 , 6) how many paths are there that reach the point (100 , 40) that also go through a point ( n, 0) for some 10 < n < 100. The best way to answer this problem is to draw a quick sketch of a path that starts from (10 , 6) and touchs the x axis before reaching the point (100 , 40). Look at the part of this path until it touches the x axis for the first time. Reflect this part of the path in the x axis. The original point (10 , 6) becomes the point (10 , 6). It follows that the number of paths that we are looking for is equal to the number of paths that start at (10 , 6) and end at the point (100 , 40) where each step is one to the right and either move up or down at each step. This is the same as moving up 68 times and down 22 times as we move to the right 90 times. Hence the number of paths is ( 90 68 ) . Remark. Given that you had not seen this problem before, I would rate this problem as very difficult. Now that you have seen the solution, I would rate it moderate. 3. In an Urn there are 3 black balls numbered 1 to 3, 3 white balls , numbered 4 to 6 and 3 red balls, numbered 7 to 9. How many different 6 digit numbers can be formed in this way by using the digits on all possible selections of 2 black balls, 2 white balls and 2 red balls. There are several ways to go about solving this problem. Here is my approach. First letways to go about solving this problem....
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 Summer '08
 KRIEGER
 Probability, Probability theory, Numerical digit, Urn B

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