Midterm 1 Solutions
Math 5010–1, Spring 2005
1. A pizzeria advertises that it offers over 1000
varieties of pizza. Suppose that you can order
any combinations of pepperoni, mushrooms,
sausage, green peppers, onions, anchovies,
salami, bacon, olives, and ground beef. Is the
restaurant’s ad correct?
possible choices. For instance, there are 24 = 16
two-by-two binary tables. They are
0
0
1
0
0
1
1
1
2. How many permutations of {1, 2, 3, 4, 5} begin
with a one and end with a five?
Solution: One way to set down the one and the
five. Then
we have to permute the rest. There
are 3! = 6 ways to accomplish that. So there
are six such possible permutations. In fact, they
are:
(1) {1, 2, 3, 4, 5}; (2) {1, 2, 4, 3, 5};
(3) {1, 3, 2, 4, 5}; (4) {1, 3, 4, 2, 5}; (5)
{1, 4, 2, 3, 5}; and (6) {1, 4, 3, 2, 5}.
So
this problem could also be solved by direct
enumeration, as was just done.
3. How many (n × m) tables of zeroes and ones
can one form?
Solution: There are (nm) entries; each hastwo
choices (zero or one). Therefore, there are 2nm 1
0
1
1
1
1
0
0
,
0
0
0
1
,
0
0
1
1
,
0
1
1
,
0
0
,
1
0
,
1
0
1
0
1
0
1
0
0
,
0
0
1
0
,
0
0
1
1
,
1
0
0
,
1
1
,
1
1
,
1
but also correct, answer is
An alternative,
∑nm
i=0
Solution: Let S denote the collection of all possible toppings, so that #(S) = 10. Evidently,
every subset of S yields a distinct toppingcombination. For instance, S itself corresponds
to the “everything” pizza, and the empty set 0/ to
“plain.” Therefore, the number of different possible pizzas we can make at this restaurant is
the same as the number of subsets of S, which
is 210 = 1024 > 1000. The restaurant’s ad is
indeed correct .
0
,
0
1
,
0
0
,
1
1
.
1
nm
i
. (Why does this equal 2nm ?)
4. Find the coefficient of x9 in the expansion of
(2 + x)12 .
Solution: Pretend you are multiplying out (2 +
x) by itself nineteen times. The question is how
many of the resulting terms lead to x9 . In order to obtain x9 you need to choose
x’s and
nine
9 ’s. Equivx
three 2’s. Therefore, there are 12
3
alently, the coefficient of x9 is times 23
12 3
12!
2 =
.
× 8 = 1760 3
3! · 9!
5. How many different messages can be sent by
using five dashes and three dots?
Solution: Make eight places, and place the
dashes and dots on them. As soon as you
know where to place the dashes (say), then
you are done. There are 85 -many such possible choices. Therefore, the number of possible
messages, made with five dashes and three dots,
is
8
8!
8·7·6 =
=
= 56 .
5
5! · 3!
3·2