This preview shows page 1. Sign up to view the full content.
Math 55,
Second Midterm Exam
SOLUTIONS
(1) Part (a) is problem # 21 (d) on page 361. The desired permutations of the seven
letters
A,B,C,D,E,F
and
G
are in bijection with the permutations of the four letters
X,Y,F
and
G
, where
X
represents the string
ABC
and
Y
represents the string
DE
. The
number of permutations of any set with four elements is 4! = 24. Hence the answer is
24
.
For part (b) we observe that there are 7! = 5040 permutations in total, and in half
of them the letter
A
precedes the letter
B
, and in the others the letter
B
precedes
A
.
Switching
A
and
B
de±nes a bijection between these two subsets. The answer is
2520
.
(2) This is problem # 25 on page 440. First consider the case
n
= 1. If one fair coin is
²ipped, then
X
1
(heads) =

1,
X
1
(tails) = 1, the expected value is
E
(
X
1
) = (1
/
2)(

1) +
(1
/
2)(1) = 0, and the variance is
V
(
X
1
) =
E
(
X
2
1
)

E
(
X
1
)
2
= 1

0 = 1. By linearity of
expectation (Theorem 3 on page 429), we have
E
(
X
n
) =
n
·
E
(
X
1
) = 0. By linearity of
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 10/10/2009 for the course MATH 55 taught by Professor Strain during the Spring '08 term at University of California, Berkeley.
 Spring '08
 STRAIN
 Math, Permutations

Click to edit the document details