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
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 Spring '08
 STRAIN
 Math, Permutations, Probability theory, Recurrence relation, Midterm Exam Solutions, random variable Xn

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