Math 55: Discrete Mathematics
UC Berkeley, Spring 2009
Solutions to Homework # 7 (due March 16, 2009)
§
5.5, #10: a) There are 6 types of croissants. This is a case of 12combinations
with repetition, or, equivalently, putting 12 indistinguishable objects
in 6 distinguishable boxes. The number of possibilitites is
(
12+6

1
12
)
=
6188.
b) Similar to above, the number of possibilities is
(
36+6

1
36
)
= 749398.
c) We can suppose that the first dozen consist of 2 of each of the 6 kinds.
Then the number of possibilities is the same as the number of ways of
choosing the second dozen, which is 6188, the same as in part a.
d) Choosing 24 croissants with exactly
k
of them broccoli is the same
as choosing 24

k
croissants among 5 kinds.
Thus, the answer is
(
24+5

1
24
)
+
(
23+5

1
23
)
+
(
22+5

1
22
)
= 52975.
e) This is the same as choosing 24

3

5 = 16 croissants. The number
of ways is
(
16+6

1
16
)
= 20349.
f) This is the same as choosing 24

1

2

3

1

2 = 15 croissants with
no more than 3 of them broccoli. Similar to part d, this is
(
15+5

1
15
)
+
(
14+5

1
14
)
+
(
13+5

1
13
)
+
(
12+5

1
12
)
= 11136.
§
5.5, #15
: a) The same as the nonnegative integers wich add to 20:
(
20+5

1
20
)
=
10626.
b) The same as the nonnegative integers adding to 11, of which there
are
(
11+5

1
11
)
= 1365.
c) Similar to part a, the number of nonnegative integers with
x
1
≥
11
is
(
10+5

1
10
)
= 1001. The number of solutions with all nonnegative in
tegers is
(
21+5

1
21
)
= 12650. Subtracting, we get 11649.
d) Similar to parts a and b, this is the same as the number of nonnega
tive integer solutions to
x
1
+
x
2
+
x
3
+
x
4
+
x
5
= 5 subject to
x
1
≤
3 and
x
2
≤
2. Without these last two conditions, the number of solutions is
(
5+5

1
5
)
= 126. Of these, the number with
x
1
≥
4 is
(
1+5

1
1
)
= 5 and
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
the number with
x
2
≥
3 is
(
2+5

1
2=15
)
. Since there are no solutions with
both
x
1
≥
4 and
x
2
≥
3, then the number of solutions with
x
1
≤
3 and
x
2
≤
2 is 126

5

15 = 106.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 STRAIN
 Math, Conditional Probability, Probability theory, Negative and nonnegative numbers, Natural number

Click to edit the document details