Unformatted text preview: (8! / (2! × 3! × 3!)) 6 8 = 0 . 000333 . 15. There are n + m drinks total. Assume that the men choose their drinks sequentially. There are n + m ways for the ±rst man to choose, n + m-1 for the second, . .., n + m-( n-1) = m + 1 for the n th. Therefore, there are ( n + m ) ··· ( m + 1) = ( n + m )! /m ! ways for the men to choose drinks. There is precisely one way that the men choose their drinks correctly. Thus, the probability of this happening is 1 / (( n + m )! /m !) = m ! / ( n + m )!. 1...
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This note was uploaded on 03/31/2008 for the course MATH 331 taught by Professor Anderson during the Spring '08 term at University of Wisconsin.
- Spring '08