Math 4653: Elementary Probability: Spring 2007
Homework #1. Problems and Solutions
1. Appendix 1 (vi):
Prove that
±
2
n
n
²
=
n
X
k
=0
±
n
k
²±
n
n

k
²
=
n
X
k
=0
±
n
k
²
2
.
Solution.
The left side is the number of all subsets of the set
{
1
,
2
, . . . , n

1
, n, n
+1
, . . . ,
2
n
}
,
which consist of
n
elements. We can count this number as follows. For each
k
= 0
,
1
, . . . , n
,
there are
(
n
k
)
possible ways to choose
k
elements from
{
1
,
2
, . . . , n

1
, n
}
, and independently,
there are
(
n
n

k
)
=
(
n
k
)
ways to choose the remaining
n

k
elements from
{
n
+ 1
, . . . ,
2
n
}
. By the
multiplication rule, there are
(
n
k
)(
n
n

k
)
ways to choose
n
elements from
{
1
,
2
, . . . , n, n
+1
, . . . ,
2
n
}
,
such that exactly
k
are taken from
{
1
,
2
, . . . , n

1
, n
}
. Finally, in order to ﬁnd the total number
with arbitrary
k
, we need to do summation with respect to
k
.
2. Appendix 1 (x):
How many diﬀerent elevenletter words (not necessarily pronounceable
or meaningful!) can be made from the letters in the word MISSISSIPPI?
Solution.
If all eleven letters were distinct, then there would be 11! ways to arrange the
letters. However, there are 4! ways to rearrange the four identical ”i”’s, another 4! ways to
rearrange the four identical ”s”’s, and 2! ways to rearrange two identical ”p”’s, which yields the
answer
11!
4!
·
4!
·
2!
= 34650
.
3.
Find the number of nonnegative integer solutions of the inequality
x
1
+
x
2
+
x
3
+
x
4
≤
4.
Hint:
Use the analogy with the number of ways to place
n
identical balls into
k
diﬀerent boxes,
which is
(
n
+
k

1
n
)
.
Solution.
The given inequality is equivalent to the equality
x
1
+
x
2
+
x
3
+
x
4
+
x
5
= 4 with
x
5
= 4

(
x
1
+
x
2
+
x
3
+
x
4
). The number of solutions is the same as the number of ways to place
n
= 4 identical balls into
k
= 5 diﬀerent boxes, which is
±
8
4
²
= 70
.
4. Sec. 1.1, #2:
Suppose a word is picked at random from this sentence. Find:
a) the chance that the word has at least 4 letters;
b) the chance that the word contains at least 2 vowels (a, e, i, o, u);
c) the chance that the word contains at least 4 letters and at least 2 vowels.
Solution.