Solutions to Homework 5, Math 55
Section 4.1
6. By the product rule, there are 4
·
6 = 24 such routes.
18.
(a) The positive integers less than 1000 which are divisible by 7 are 7
·
1
,
7
·
2
, . . . ,
7
·
999
7
. Therefore,
999
7
= 142 numbers less than 1000 are divisible by 7.
(b) The numbers less than 1000 which are divisible by both 7 and 11 are the ones divisible by 77; therefore,
there are
999
77
= 12 of these. Subtracting, 142

12 = 130 of the numbers divisible by 7 are not divisible
by 11.
(c) By the previous part, 12 are.
(d) There are
999
11
= 90 multiples of 11 less than 1000. Therefore, by inclusionexclusion, 142+90

12 =
220 are divisible by either 7 or 11.
(e) By part (b), there are 130 multiples of 7 not divisible by 11. Similarly, there are 90

12 = 78 multiples
of 11 not divisible by 7. Therefore, 130 + 78 = 208 are divisible by exactly one of 7 and 11.
(f) Since 220 are divisible by either 7 or 11 by part (d), 999

220 = 779 are divisible by neither 7 nor 11.
(g) There are 9 such 1digit numbers. For a 2digit number, we can choose the first digit to be 1 through
9, then we have 9 choices for the second digit; thus, there are 9
·
9 = 81 such 2digit numbers. Similarly,
there are 9
·
9
·
8 = 648 such 3digit numbers. In total, there are 9 + 81 + 648 = 738 such numbers less
than 1000.
(h) There are 4 such 1digit numbers. For a 2digit number, if we choose one of the 4 possible even digits
for the first digit, there are 4 possibilities for the second digit; if we choose one of the 5 possible odd
digits for the first digit, there are 5 possibilities for the second digit. Therefore, there are 4
·
4+5
·
5 = 41
such 2digit numbers.
Similarly, for a threedigit number, if the first digit is even (4 choices), then
there are 4 choices for the third digit, then 8 choices for the second digit. If the first digit is odd (5
choices), then there are 5 choices for the third digit, then 8 choices for the second digit. Therefore,
there are 4
·
4
·
8 + 5
·
5
·
8 = 328 such 3digit numbers. In total, there are 4 + 41 + 328 = 373 such
numbers less than 1000.
22.
(a) Since there are 10 ways to choose the first digit, then 9 ways to choose the second, and so on, there
are 10
·
9
·
8
·
7 = 5040 strings of four decimal digits which do not contain the same digit twice.
(b) There are 10 ways to choose each of the first three digits, and 5 ways to choose the last digit as even.
Therefore, there are 10
3
·
5 = 5000 such strings.
(c) There are 4 ways to choose which digit is
not
a 9, then 9 ways to choose that digit. Therefore, there
are 4
·
9 = 36 such strings.
40. Out of these bit strings, 2
5
= 32 start with two 0’s; 2
4
= 16 end with three 1’s; and 2
2
= 4 start with two
0’s and end with three 1’s. Therefore, by inclusionexclusion, 2
5
+ 2
4

2
2
= 44 start with two 0’s or end
with three 1’s.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 STRAIN
 Math, ways

Click to edit the document details