HW10 Solutions
Dilip Raghavan
November 23, 2008
1
Section 8.2
Problem 12.
Let
U
be the set of all 4–digit numbers, including those with leading 0s. For each
i
between
0 and 9, let
A
i
be the set of all 4–digit numbers in which the digit
i
occurs exactly twice. Now,
U
=
10
4
.
For each
i
,

A
i

=
4
2
×
9
2
– first choose the 2 slots for the exactly 2 occurances of
i
, and then arrange any
of the other 9 digits in the two remaining slots. Therefore,
S
1
=
10
1
×
4
2
×
9
2
. Now, given a pair
i
j
,
A
i
∩
A
j
=
4
2
– choose 2 slots for the two occurrences of
i
, and then
j
must go into the remaining 2 slots.
Therefore,
S
2
=
10
2
×
4
2
. Notice that all triple intersections are empty because a 4–digit number cannot
have 3 digits ocurring exactly twice. Thus,
¯
A
0
∩ · · · ∩
¯
A
9
=
U
S
1
+
S
2
=
10
4

10
1
×
4
2
×
9
2
+
10
2
×
4
2
.
Problem 18.
For a real number
x
, let
x
denote the least integer
greater than or equal
to
x
. Note that the
number of odd numbers in the set
{
1
, . . . ,
n
}
is equal to
n
2
. Now, let
U
be the set of all possible arrangements
of 1
,
2
, . . . ,
n
. For each odd number
i
between 1 and
n
, let
A
i
be the set of arrangements with
i
ocurring in
slot
i
(so
i
is
not
deranged).
U
=
n
!. For each odd number
i
,

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 Spring '11
 Hernandez/Rag
 Elementary arithmetic, Zagreb

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