Download Math 316 – Solutions To Sample Exam 1 Problems

Solutions to Sample Exam 1 Problems – Math 316
1
Math 316 – Solutions To Sample Exam 1 Problems
1. A bricklayer is laying a row of 20 bricks. In how many ways can she do this if she has:
(a) plentiful red bricks and plentiful white bricks?
(b) 13 red bricks and 7 white bricks?
(c) 20 red bricks and 17 white bricks?
Solution:
(a) There are 20 positions that we want to fill with bricks, and for each position, there are two choices: red
or white. Therefore, our answer is 220 .
(b) We break the bricklaying process into 2 tasks:
20
• Lay down the white bricks ←−
ways
7
13
• Lay down the red bricks ←−
= 1 way
13
20
13
20
Therefore, our answer is
·
=
. Note that, if we had laid the red bricks first, we would
7
13
7
20
, which is also the correct answer.
have obtained
13
(c) Observe that our answer should be the answer to (a), minus the number of arrangements that contain 18
or more white bricks. Using the same logic as in part (b) above, there are C(20, 18) arrangements that
contain 18 white bricks, C(20, 19) that contain 19 white bricks, and C(20, 20) = 1 arrangement that
contains 20 white bricks. Our answer is therefore
20
20
20
− 1.
−
2 −
19
18
2. In a soccer game, suppose that each team scores 5 or fewer goals. How many different final scores are possible
in each of the following situations?
(a) The game cannot end in a tie.
(b) The game might end in a tie.
(c) Two specific teams, Sonoma State and Chico State are playing, and the game cannot end in a tie.
(d) Sonoma State and Chico State are playing, and the game might end in a tie.
Solution:
(a) Since the teams are not specified, it makes sense to consider 5 to 1 the same final score as 1 to 5; in other
words, the order of the two scores is irrelevant. Therefore, the number of final scores is equivalent
to the
number of ways to choose 2 numbers from the set {0, 1, 2, 3, 4, 5} , giving us a final answer of 62 = 15.
(b) Our answer should be the answer
to (a), plus the 6 possible tie scores: 0 − 0, 1 − 1, 2 − 2, . . . , or 5 − 5.
Therefore, our answer is 62 + 6 = 21.
(c) We break the task of choosing a final score into two parts: (1) choose a score for Sonoma State in 6
ways, and (2) choose a final score for Chico State, different from SSU’s final score, in 5 ways. Our
answer is therefore 6 · 5 = 30.
(d) This time, we can choose a final score for SSU in 6 ways, but since ties are allowed, there are still 6
possible choices of scores for Chico State. Therefore, our answer is 62 = 36.
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Solutions to Sample Exam 1 Problems – Math 316
3. Suppose we have 27 distinct positive odd integers, all less than 100. Show that there is at least one pair
of our 27 integers whose sum is 102. (Hint: Use {3, 99} , {5, 97} , {7, 95} , . . . , {49, 53} , {1} , {51} as your
pigeonholes.)
Solution:
First, we note that the list of all positive odd integers less than 100 looks like
1, 3, 5, 7, . . . , 99, and note that there are 50 numbers on this list. We also observe that 48 of the 50 numbers
form pairs that add up to 102, so we designate our pigeonholes as follows:
{3, 99} , {5, 97} , {7, 95} , . . . , {49, 53} , {1} , {51}
Thus, we have 24 pigeonholes that consist of 2-element sets, and 2 pigeonholes that consist of 1-element sets,
giving a total of 26 pigeonholes. Regarding our 27 numbers as pigeons, we conclude that there must be at
least two integers in one of our sets. Since there cannot be 2 numbers in either of the 1-element sets, this
means that one of the 24 sets {3, 99} , {5, 97} , . . . , {49, 53} must contain two of our numbers. Therefore,
whatever these two numbers are, they must add up to 102.
4. Suppose we know that Safeway stocks 20 different types of sweetened cereal.
(a) Assuming there are lots of each type, in how many ways can we choose four boxes of sweetened cereal
to buy?
(b) When we get there, it turns out that the store has lots of Lucky Charms but only one each of the other
19 kinds of sweetened cereals. In how many ways can we choose four boxes of sweetened cereal now?
(c) Suppose we want to choose 8 boxes of cereal, including at least 2 boxes of Honey Smacks. Assuming
lots of each kind are again available, how many choices are there?
Solution:
(a) Viewing the 20 kinds of cereal as 20 categories, our answer
is the number of different star/bar sequences with 4 stars
(representing the 4 boxes of cereal we choose) and 19 dividing bars to separate the 20 categories. (Two such star/bar
sequences are shown to the right). This gives 19 + 4 = 23
positions to fill with stars and bars, and there are C(23, 4)
ways to place the stars, and once the stars are placed, only
one way to place the bars. Therefore, our answer is 23
4 .
** *
*** *
*
Figure 1. The top sequence represents
choosing 2 cereal boxes of type 1, 1 box
of type 2, and 1 box of type 20. The bottom sequence represents choosing 3 boxes
of type 5 and 1 box of type 6.
(b) We divide the problem into different cases depending on how many boxes of Lucky Charms we choose.
Case 1. (no boxes of Lucky Charms)
• Choose no boxes of Lucky Charms ←− 1 way
• Choose 4 boxes of cereal from remaining 19 kinds ←− C(19, 4) ways
(Since there’s only one type of each of the remaining 19 types of cereals, repetition is not allowed.)
Case 2. (1 box of Lucky Charms)
• Choose 1 box of Lucky Charms ←− 1 way
• Choose 3 boxes of cereal from remaining 19 kinds
←−
C(19, 3) ways
Similarly, Cases 3, 4, and 5 yield C(19, 2), C(19, 1), and C(19, 0) choices. Our final answer is therefore
19
19
19
19
19
+
+
+
+
= 5036.
4
3
2
1
0
(c) Since we know that we want to have at least 2 boxes of Honey Smacks, we really only need to count the
number of ways to choose the remaining 6 boxes of cereal. Using the same logic of part (a), we have 6
stars and 19 bars, giving an answer of 25
6 .
Solutions to Sample Exam 1 Problems – Math 316
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5. The SSU Academic Senate has 35 elected members.
(a) How many different subcommittees can be formed, assuming that each subcommittee must have at
least 1 member?
(b) How many different subcommittees can be formed that have exactly 5 members?
(c) How many different subcommittees are there that contain Senator Cabaniss as a member?
Solution:
(a) Choosing a subcommittee is equivalent to choosing a nonempty subset of the 35 elected members.
Therefore, our answer is 235 − 1.
(b) This is equivalentto counting the number of subsets a 35-element set that have exactly 5 elements, so
our answer is 35
5 .
(c) Note that every subcommittee that contains Senator Cabaniss as a member consists of Senator
Cabaniss, plus some (possibly empty) subset of the remaining 34 elected members of the senate.
Therefore, our answer is 234 .
6. In how many ways are there to pick a collection of exactly 20 jelly beans from a vat that contains many
peach, mango, kiwi, coconut, and licorice jelly beans if
(a) there must be at least 10 peach jelly beans?
(b) there must be at most 10 peach jelly beans?
Solution:
(a) We break this task into two parts:
←−
only 1 way
• Choose 10 peach jelly beans
• Choose 10 more jelly beans of any type
←−
C(14, 10) ways
(Since we can choose the remaining 10 jelly beans from any of the 5 flavors, we have 10 stars and 4
bars, giving C(14, 10) choices. One possible star/bar sequence is illustrated below:)
peach
mango
kiwi
coconut
licorice
∗∗ |
∗∗
|
∗
|
∗∗∗
| ∗∗
14
Therefore, our answer is
= 1001.
10
(b) Our strategy will be to first count the total number of ways to choose 20 jelly beans from the vat with
no restrictions, and then subtract the jelly bean choices that contain at least 11 peach jelly beans. In
counting all possible choices of 20 jelly beans, note that we have 20 stars and 4 bars, giving C(24, 20)
choices. In counting those choices that contain at least 11 peach jelly beans, we use a similar argument
to part (a) above, obtaining C(13, 9) choices. Our answer is therefore
24
13
−
= 9911.
20
9
7. (a) By substituting appropriate values into the binomial theorem, find a formula for the sum
n
n
n n
3 n
2 n
.
+ ··· + 3
+3
+3
+3
n
3
2
0
1
(b) Give a combinatorial proof of the formula you found in part (a). (Hint: Count the number of ways to
paint n different houses either red, blue, green, or not at all.)
Solution:
Solutions to Sample Exam 1 Problems – Math 316
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(a) Letting x = 1 and y = 3 in the binomial theorem, we have
n n
n n−1
n n−2 2
n n−3 3
n 0 n
n
n
4 = (1 + 3)
=
1 +
1
·3 +
1
·3 +
1
·3 + ··· +
1 ·3
0
1
2
3
n
n
n
n n
3 n
2 n
,
=
+ ··· + 3
+3
+3
+3
n
3
2
0
1
so our formula for the sum is 4n .
(b) The formula that we are trying to prove is
n
n
n
n
n
= 4n .
+ · · · + 3n
(∗)
+ 33
+3
+ 32
n
3
2
0
1
We claim that both sides of (∗) count the number of ways to paint n distinct houses either red, blue,
green or not at all, which we verify below:
Left-hand side: Here, we enumerate the number of different ways to paint by using cases according to the
number of houses we paint.
Case 1. (No houses get painted.) In this case, we are effectively choosing 0 houses to paint, which can be
done in C(n, 0) ways.
Case 2. (1 house gets painted.) We break this task into two parts: pick 1 house to be painted in C(n, 1)
ways, and then choose to paint it either red, green, or blue in 3 ways.
Case 3. (2 houses get painted.) Again, break this task into two parts: pick 2 houses to be painted in
C(n, 2) ways, and then choose colors for both houses in 32 ways.
..
.
Case n. (n houses get painted.) Pick all n houses to be painted in C(n, n) ways, and then choose colors
for all n houses in 3n ways.
Summing over all the cases above accounts for all ways of painting and gives the left-hand side of (∗).
Right-hand side: Here, we enumerate the number of different ways to paint by simply observing that for
each house, we have four choices: either paint it red, blue, green, or don’t paint it at all. Therefore, there
are 4n total ways to paint, which gives the right-hand side of (∗).