This preview shows pages 1–2. Sign up to view the full content.
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!)
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '10
 RohiniKumar

Click to edit the document details