Math 55: Discrete Mathematics, Fall 2008
Homework 8 Solutions
*[3,3 and 4 points] 5.5: 10(a)
(
12+6

1
12
)
(choosing 12 things from 6 with repetitions)
(c)
(
12+6

1
12
)
(the ﬁrst two of each kind make 12 and you have a dozen left to choose freely)
(f)
(
15+6

1
15
)

(
11+6

1
11
)
(count choices without any restriction on broccoli, then subtract
those with at least 4 broccoli)
34.
(
5
3
,
2
)
+
(
5
3
,
1
,
1
)
+
(
5
2
,
3
)
+
(
5
2
,
2
,
1
)
+
(
5
1
,
3
,
1
)
+
(
6
3
,
3
)
+
(
6
3
,
2
,
1
)
+
(
6
2
,
3
,
1
)
+
(
7
3
,
3
,
1
)
(one term for
each possible number of E’s, S’s and R’s adding up to at least 5)
42.
(
52
13
,
13
,
13
,
13
)
= 52!
/
(13!)
4
6.1: 8.
(
51
4
)
/
(
52
5
)
16. 4
(
13
5
)
/
(
52
5
)
18. 4
·
10
/
(
52
5
)
30. 6
·
34
/
(
40
6
)
(I assume that the problem intends an
unordered
subset of 6 integers
chosen at random from
{
1
,...,
40
}
. The numerator counts subsets that match exactly 5 of
the player’s chosen 6.)
36. 8 is more likely with two dice than three. With two dice,