Download ExamView - Chapter 4 and 5 Test.tst

ID: A
Unit 1 Test Organized Counting
Answer Section
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MULTIPLE CHOICE
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Section 4.1
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Section 4.2
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Section 4.2
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Section 4.2
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Section 5.1
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Section 5.1
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Section 5.1
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Section 5.2
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Section 5.3
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Section 5.3
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Section 5.4
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Section 5.4
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Section 4.1
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Section 4.2
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Section 4.2
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1
REF: Knowledge & Understanding
Fundamental counting principle
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REF: Knowledge & Understanding
Factorials
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REF: Knowledge & Understanding
Permutations
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REF: Knowledge & Understanding
Permutations
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REF: Knowledge & Understanding
Organized counting
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REF: Knowledge & Understanding
Organized counting
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REF: Knowledge & Understanding
Organized counting
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REF: Knowledge & Understanding
Evaluating combinations
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REF: Knowledge & Understanding
Applying combinations
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REF: Knowledge & Understanding
Applying combinations
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REF: Knowledge & Understanding
Binomial theorem
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REF: Knowledge & Understanding
Binomial theorem
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REF: Knowledge & Understanding
Fundamental counting principle
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REF: Knowledge & Understanding
Permutations
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REF: Knowledge & Understanding
Permutations
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ID: A
SHORT ANSWER
16. ANS:
There are 12 possible meals at this Chinese restaurant.
PTS: 1
REF: Applications OBJ: Section 4.1
17. ANS:
There are 7 200 000 000 telephone numbers possible.
TOP: Tree diagrams
PTS: 1
18. ANS:
9
REF: Applications OBJ: Section 4.1
TOP: Fundamental counting principle
PTS: 1
19. ANS:
24
REF: Applications OBJ: Section 4.2
TOP: Permutations
PTS: 1
REF: Applications OBJ: Section 4.2
20. ANS:
There are 57 subcommittees that can be formed.
TOP: Permutations
PTS: 1
REF: Applications OBJ: Section 5.3 TOP: Applying combinations
21. ANS:
The first three terms are
1(2x)7 = 128x7
7(2x)6(–y) = – 448x6y
21(2x)5(–y)2 = 672x5y2
PTS: 1
REF: Knowledge & Understanding
TOP: Binomial theorem
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OBJ: Section 5.4
ID: A
PROBLEM
22. ANS:
Steve has four choices for the first question, but only three choices for each of the remaining questions since
he does not choose answers with the same letter twice in a row. From the multiplicative counting principle,
there are 4 × 3 × 3 × 3 × 3 = 324 ways Steve can answer the five questions.
PTS: 1
REF: Thinking/Inquiry/Problem Solving | Communication
OBJ: Section 4.1 TOP: Fundamental counting principle
23. ANS:
The number must end in 1, 3, 5, 7, or 9, so there are five choices for the last digit. There are only eight
choices for the first digit since it cannot be either 0 or the same as the last digit. There are eight choices for
the second digit since it cannot be the same as the first or last digits. Similarly, there are seven choices for the
third digit and six choices for the fourth digit. Using the multiplicative counting principle, there are
8 × 8 × 7 × 6 × 5 = 13 440 five-digit odd numbers with no repeated digits.
PTS: 1
REF: Communication | Thinking/Inquiry/Problem Solving
OBJ: Section 4.1 TOP: Fundamental counting principle
24. ANS:
a) 7P7 = 5040
b) 6P6 = 720
c) Treat r and s as a unit. This pair can be arranged in 6P6 ways with the remaining letters. The pair itself can
be arranged as rs or sr, so there is a total of 6P6 × 2 = 1440 arrangements with r and s adjacent. Subtract
5040-1440 = 3600
PTS: 1
REF: Thinking/Inquiry/Problem Solving OBJ: Section 4.2
TOP: Permutations
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ID: A
25. ANS:
Let n be the number of students who take both geometry and mathematics of data management. Applying the
principle of inclusion and exclusion,
125 = 64 + 40 + 51 – 12 – 11 – n + 8
n = 15
This number includes the students who take all three courses, so the number who take just geometry and
mathematics of data management is 15 – 8 = 7.
PTS: 1
REF: Applications OBJ: Section 5.1 TOP: Venn diagrams
26. ANS:
Direct Method
There are 6 pieces for piano and 12 for the other instruments. The students can choose 1, 2, 3, or 4 piano
pieces. Consider each of these cases in turn.
1 piano piece: The students can choose the piano piece in C(6, 1) ways and the remaining 3 pieces in C(12, 3)
ways. The number of combinations with 1 piano piece is C(6, 1) × C(12, 3) = 1320.
2 piano pieces: The number of combinations is C(6, 2) × C(12, 2) = 990.
3 piano pieces: The number of combinations is C(6, 3) × C(12, 1) = 240.
4 piano pieces: The number of combinations is C(6, 4) × C(12, 0) = 15.
The number of combinations that include at least 1 piano piece is the total of these four cases, 2565.
Indirect Method
Find the total number of possible combinations and subtract those that do not have any piano pieces:
C(18, 4) – C(12, 4) = 3060 – 495
= 2565
PTS: 5
REF: Applications | Communication
TOP: Applying combinations
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OBJ: Section 5.3