Solution Ideas for 201001 Math 222 Practice Midterm 1
Questions
1.
(a) If the first digit is 7, there are
(
4
2
)
9
2
such integers, and if the first
digit isn’t 7 there are
(
4
3
)
9
·
8 of them. Hence, by the AP, the total
number is
(
4
2
)
9
2
+
(
4
3
)
9
·
8.
(b) If the first digit is even, then the number of such integers is 4(
(
4
2
)
5
4
+
(
4
3
)
5
4
+
(
4
4
)
5
4
), and if the first digit is odd, there are 5(
(
4
3
)
5
4
+
(
4
4
)
5
4
)
of them.
Hence, by the AP, the total number is 4
·
5
4
(
(
4
2
)
+
(
4
3
)
+
(
4
4
)
) + 5
5
(
(
4
3
)
+
(
4
4
)
).
(c) 9
·
9
·
8
·
7
·
6
2. There are 61
choose
2 choices for the pair (
x
2
, x
3
) – once the two numbers
are known the one encountered first when travelling clockwise from 0 must
be
x
2
. There are then 60 choices for
x
1
6
=
x
2
. Hence the number of valid
combinations is 60
(
61
2
)
3. 2
15

2
7
.
4.
(a) The integer
a
∈ {
1
,
2
, . . . ,
10
}
belongs to the subset
S
f
if and only if
f
(
a
) = 1.
(b) If
f
and
g
are different functions, then
f
(
a
)
6
=
g
(
a
) for some
a
∈
{
1
,
2
, . . . ,
10
}
. Therefore
a
belongs to one of
S
f
and
S
g
(see (a) for
notation), but does not belong to the other one.
(c) Given
S
⊆ {
1
,
2
, . . . ,
10
}
, define
f
by setting
f
(
a
) = 1 when
a
∈
S
and
f
(
a
) = 0 otherwise.
5.
(a) 12!
/
6.
(b) Seat Gary on a fixed side of the table (2 ways) and seat everybody
else relative to him. There are 5! ways to assign the remaining men
to the remaining 5 sides of the table, and each of these men has 2
choices of seat. This leaves one seat on each side of the table to be
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 Spring '11
 HuangJing
 Division, Integers, AP, Numerical digit, Natural number

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