EECS 203: Homework 7 Solutions
Section 5.1
1. (E) 24
a) Since the strings do not contain the same digit, there are 10 ways to choose the first digit, 9 ways to
choose the second and so on. Therefore the answer is
10
·
9
·
8
·
7=5040
.
b) There are 10 ways to choose each of the first three digits and 5 ways to choose the last digit. So the
answer is
10
·
10
·
10
·
5=5000
.
c) There are 4 ways to choose the position that is not 9 and 9 ways to choose the digit in that position.
Therefore the answer is
4
·
9=36
.
2. (E) 30fgh
f) The first two letters are fixed and we can choose 6 letters for the remaining 6 positions. Hence the
answer is
26
6
=308915776
.
g) The first and last two letters are fixed and we can choose 4 letters for the remaining 4 positions. Thus
the answer is
26
4
=456976
.
h) From part f, we have
26
6
strings that start with the letters BO. Similarly, it can be shown that there
are
26
6
strings that end with the letters BO. By part g, we know that there are
26
4
strings that start
and end with BO. So, by the inclusion exclusion principle, there are
26
6
+26
6
−
26
4
strings that start
or end with BO. The answer is
26
6
+26
6
−
26
4
=617374576
.
3. (E) 34
Each element in the domain
{
1
,
2
,...,n
}
can map to either
0
or
1
. Thus there are
2
n
distinct functions
from the set
{
1
,
2
,...,n
}
to the set
{
0
,
1
}
. The answer is
2
n
.
4. (E) 40
a) We first need to choose 5 more people from the group except the bride and there are
C
(9
,
5)
such
ways. Moreover there are
6!
ways to arrange the selected 5 people and the bride in a row. Thus, the
answer is
C
(9
,
5)
·
6!=90720
.
b) We need to choose 4 more people except both the bride and the groom, giving us
C
(8
,
4)
ways. Since
we have
6!
ways to arrange 6 people in a row, the answer is
C
(8
,
4)
·
6!=50400
.
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 Spring '07
 YaoyunShi
 Numerical digit, consecutive 0s, Groom, Texas

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