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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 Spring '10
 RohiniKumar
 Blackandwhite films, Mississippi, #Total, #with, #without

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