MDM4U Grade 12
Permutations and Organized Counting Test 2
Multiple Choice
Identify the letter of the choice that best completes the statement or
answers the question.
1. A card is selected from a standard deck. In how many ways could it
be a face card or a 10?
a. 12 + 4
b. 12 – 4
c. 12 × 4
d. 12
Answer: A
2. John owns a small Internet company and offers his clients three
different formats for the name part and seven different domains for the
second part of their e-mail address. In how many ways can a person set
up an e-mail address through John’s company?
a. 10
b. 21
c. 4
d. 12
Answer: B 3 x 7 = 21
3. At the cafeteria, Jeff has a choice of three soups, four types of
sandwich, and two desserts. In how many ways can Jeff choose his
lunch if he has soup, a sandwich, and dessert?
a. 9
b. 18
c. 24
d. 12
Answer: C 3 x 4 x 2 = 24.
4. The final score of a hockey game is 4 to 3. How many different scores
could there have been at the end of the first period?
a. 7
b. 12
c. 20
d. 9
Answer: C (4 + 1) x (3 + 1) = 20.
5. How many different sums of money can you make with three coins of
different denominations?
a. 23
b. 3
c. 23 – 1
d. 32
Answer: C
6. (n + 1)(n + 2)n! can be simplified to
a. (n2 + 2)n!
b. (n2 + 4n + 2)!
c. (n + 2)!
Answer: C
7. The value of 12P2 is
a. 12
b. 132
c. 20 734
Answer: B
d. (n2)! + 2!
d. 479 001 600
8 Which expression cannot be evaluated?
98!
100!
a. 100P98
b.
c. P (98, 100)
d.
100!
98!
Answer: C
9. Find the number of paths from A to B, travelling along the lines,
downward or to the right.
a. 252
b. 462
Answer: B
c. 310.
d. 550
10. Given sets A = {t, u, v, w, x, y, z} and B = {p, q, r, s, t, u}, A∩B =
a. {p, q, r, s, v, w, x, y, z}
c. {t, u}
b. {p, q, r, s, t, u, v, w, x, y, z}
d. {p, q, r, s, t, t, u, u, v, w, x, y, z}
Answer: C
Short Answer
1. How many ways are there to draw a 7 or a king from a standard deck
of 52 playing cards?
Answer: There are 4 + 4 = 8 ways to draw a 7 or a king.
2. In how many ways can you roll a sum of 6 or a sum of 10 with a pair
of dice?
Answer: You can roll a sum of 6 or a sum of 10 in 5 + 3 = 8 ways.
3. A Chinese restaurant features a lunch special with a choice of wonton
soup or spring roll to start, sweet and sour chicken balls, pork, or beef
for the main dish, and steamed or fried rice as a side dish. Create a tree
diagram to show all the possible lunch specials at this restaurant. How
many different possibilities are there?
Answer:
There are 12 possible meals at this Chinese restaurant.
4. On his university application, Ervis must list his course choices in
order of preference. He must choose three of the four courses available
in his major discipline, and two of the three courses offered in related
subjects. In how many ways can Ervis list his course choices? Explain
your reasoning.
Answer:
Ervis has 144 ways to choose his courses since he can choose his major
courses in 4 × 3 × 2 ways and his options in 3 × 2 ways. By the
multiplicative rule, he has 24 × 6 = 144 ways in which he can rank the
courses.
5. Bill works in an ice cream store for the summer months. How many
different two- or three-scoop cones can Bill create if he has chocolate,
mint chocolate chip, vanilla, maple walnut, and pistachio ice cream
available?
Answer: There are 5 x 4 + 5 x 4 x 3 = 80 ways in which Bill can create
two- or three-scoop cones.
6. In how many ways can a student answer all of the questions on a
true/false test that has eight questions? Explain your solution.
Answer: For each question there are two choices. Applying the
multiplicative (fundamental) counting principle, there are 28 = 256 ways
a student could answer the test.
7. In how many ways can the interviewers select a first, second, and
third choice from a group of seven applicants for a position at a law
firm?
Answer: There are P (7, 3) = 210 ways to select the top three choices.
8. How many ten-digit telephone numbers are possible if the first three
digits must all be different?
Answer: There are 10 x 9 x 8 x 107 = 7 200 000 000 telephone numbers
possible.
9 How many arrangements of five letters from the word certain begin
with a vowel?
Answer: There are 3 x P (6, 4) = 1080 arrangements that begin with a
vowel.
10. Simplify:
a) 12 × 11! = 12!
Answer:
(𝑛+ 3)!
b)
= n (n + 1)( n + 2)(n + 3)
(𝑛− 1)!
Answer:
12!
12!
c)
+
8! × 4!
9! × 3!
Answer: 715
d) n [n! + (n – 1)!]
Answer: n [n! + (n – 1)!] = n [n (n – 1)! + (n – 1)!]
= n (n – 1)! (n + 1)
= n! ∙ (n + 1)
= (n + 1)!
Problems
1. Steve is taking a multiple-choice test consisting of five questions that
each have four answers labelled A, B, C, and D. How many ways can
Steve answer all five questions if he does not choose answers with the
same letter for any two consecutive questions? Explain your reasoning.
Answer: 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.
2. How many arrangements of five letters from the word certain contain
the letter t? Explain your reasoning.
Solution
Direct method
The letter t can be any of the five letters in the arrangement. If t is the
first letter, there are six choices left for the second letter, five choices for
the third letter, four for the fourth, and three for the fifth. Using the
multiplicative counting principle, there are 6 × 5 × 4 × 3 = 360
arrangements with t as the first letter. There is the same number of
arrangements with the letter t in each of the other four positions. Thus,
the total number of five-letter arrangements that include the letter t is 5
× 360 = 1800.
Indirect method
Find the total number of five-letter arrangements and subtract those that
do not contain the letter t. The total number of five-letter arrangements is
7 × 6 × 5 × 4 × 3 = 2520. In arrangements without the letter t, there are
six choices for the first letter. For the second letter, there are five choices
left; for the third, four; for the fourth, three; and for the fifth, two.
Applying the multiplicative counting principle, the number of
arrangements without the letter t is 6 × 5 × 4 × 3 × 2 = 720. Therefore,
the number of arrangements with the letter t is 2520 – 720 = 1800.
3. How many odd five-digit numbers have all the digits different?
Explain how you arrived at your answer.
Answer: 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.
4. In Switzerland the postal codes consist of two letters followed by four
numbers.
a) How many postal codes are possible in the Swiss system if there are
no restrictions on the choice of letters or numbers?
Answer: There are 26 x 26 x 104 = 6 760 000 postal codes possible.
b) How many postal codes are possible in the Swiss system if you
cannot use the letter O and the number 0?
Answer: There are 4 100 625 postal codes possible.
n (no O and 0) = 25 x 25 x 94 = 4 100 625
c) How many postal codes are possible in the Swiss system if you cannot
use the number 0 or the number 0?
Answer: There are 6 584 611 postal codes possible.
n (no O or 0) = n (no O) + n (no 0) – n (no O and 0)
= 25 x 25 x 104 + 26 x 26 x 94 – 25 x 25 x 94 = 6 584 611
5. Find the number of divisors, other than one, of 2700?
Answer: 2700 = 22 × 33 ×52 ∴
(2 + 1) × (3 + 1) × (2 + 1) – 1 = 3 × 4 × 3 – 1 = 35.
6. In a survey of 56 Mathematics students who wrote the Advanced
Functions, Calculus and Data Management:
33 passed Advanced Functions
26 passed Calculus
24 passed Data Management
10 passed Advanced Functions and Calculus
8 passed Advanced Functions and Data Management
14 passed Calculus and Data Management
4 passed all three exams
a) Illustrate this information on a Venn diagram.
Answer:
b) How many students passed Calculus but not Data Management?
Answer: 12
c) How many students failed all the examinations?
Answer: 1
7. How many different 8-letter words can be formed using letters from
the word CANADIAN so that D always immediately precedes an A?
7!
Answer:
= 1 260
2! × 2!
8. In how many different ways could you stack 4 quarters, 3 dimes, 2
nickels and one penny?
10!
Answer:
= 12600
4! ×3! ×2!
9. How many different paths will spell the word binomial in the
following diagram?
B
I
I
N
N
N
O
O
M
I
I
A
A
A
L
L
L
Answer: 18 + 18 + 6 = 42 paths
1
1
1
1
2
1
3
3
6
6
6
6
12
6
18
18
6