Solutions from a student (dragged) 11

Solutions from a student (dragged) 11 - We have 52! unique...

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From the basic principle of counting then we have 6 · 3=18tota lpaths . Problem 23 (assignments to beds) Assuming that twins sleeping in di±erent bed in the same room counts as a di±erent arraign- ment, we have (2!) · (2!) · (2!) = 8 possible assignments of each set of twins to a room. Since there are 3! ways to assign the pair of twins to individual rooms we have 6 · 8=48poss ib le assignments. Problem 24 (practice with the binomial expansion) This is given by (3 x 2 + y ) 5 = 5 ± k =0 ² 5 k ³ (3 x 2 ) k y 5
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Unformatted text preview: We have 52! unique permutations, but since the dierent arraignments of cards within a given hand do not matter we have 52! (13!) 4 , possible bridge hands. Problem 26 (practice with the multinomial expansion) This is given by the multinomial expansion ( x 1 + 2 x 2 + 3 x 3 ) 4 = n 1 + n 2 + n 3 =4 4 n 1 , n 2 , n 3 x n 1 1 (2 x 2 ) n 2 (3 x 3 ) n 3 The number of terms in the above summation is given by 4 + 3-1 3-1 = 6 2 = 6 5 2 = 15 ....
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This note was uploaded on 02/25/2011 for the course STAT 418 taught by Professor G.jogeshbabu during the Winter '08 term at Pennsylvania State University, University Park.

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