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HW1_Solutions

# HW1_Solutions - PSTAT 120A Spring 2010 HW 1 Solutions...

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PSTAT 120A Spring 2010 HW 1 Solutions 1. (Ross 1.5) There are 8 choices for the ﬁrst digit, for each of these there are 2 choices for the second digit, and for each of these 9 choices for the last digit. Since “for each” indicates a multiplication, there are (8)(2)(9) = 144 total possible area codes under these constraints. For area codes starting with a “4”, there is only 1 choice for the ﬁrst digit, so there are (1)(2)(9) = 18 possibilities. 2. (Ross 1.8) We are asked to consider the number of diﬀerent letter arrangements possible for several words. If a word of n letters has no repeated letters, then there are n ! ways to arrange these. When letters are repeated, n ! is an overcount because it counts permuting repeated letters amongst them- selves, e.g. swapping the “p’s” in “propose”, as creating a new arrangement. To correct for this, divide by the number of ways to permute groups of repeated letters amongst themselves. (a) Fluke There are 5 letters, none of which are repeated. So there are 5! = 120 arrangements. (b) Propose There are 7 letters, with a pair of “p’s” and a pair of “o’s”. These pairs prompt a division by 2! each, to correct for the overcount mentioned above. So there are 7! (2!)(2!)

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HW1_Solutions - PSTAT 120A Spring 2010 HW 1 Solutions...

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