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Unformatted text preview: Solutions to Homework 03 1. a. For the set, we have a choice of k 1 objects from a total of n objects. Then the number of choices are N 1 = n k 1 ! since we are sampling without replacement and without ordering. For the second set B 2 , we are left with n k 1 objects to choose from. The number of ways to choose k 2 objects from here is then given by N 2 = n k 1 k 2 ! For the third set B 3 , we choose k 3 objects out of a total of the remaining n k 1 k 2 objects. The number of ways of doing this is N 3 = n k 1 k 2 k 3 ! Continuing like this, for the set B m 1 we are left with n k 1 k 2··· k m 2 objects from which we choose k m 1 objects. The number of ways of doing this is then N m 1 = n k 1 k 2 ··· k m 2 k m 1 ! For the last set B m , we have n k 1 ···  k m 1 objects from which we choose k m objects. But note that n = k 1 + k 2 + ··· k m 1 + k m ⇒ k m = n k 1 k 2 ···  k m 1 Then the number of choices N m = n k 1 ···  k m 1 k m ! = k m k m ! = k m ! k m !0! = 1 b. The number of possible outcomes for B 1 is N 1 , for B 2 is N 2 etc. Hence, the total number of possible outcomes is N 1 N 2 ··· N m = n k 1 ! n k 1 k 2 ! n k 1 k 2 k 3 ! ··· n k 1 k 2 ··· k m 2 k m 1 ! 1 = n ! ( n k 1 )! k 1 !...
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This note was uploaded on 09/07/2009 for the course ECE 302 taught by Professor Gelfand during the Fall '08 term at Purdue.
 Fall '08
 GELFAND

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