MATH 109
Counting Techniques
A fundamental practice in mathematics is determining the number of ways a certain
outcome may occur. For example, when you roll two dice, how many ways are there to
obtain a sum of seven? When you deal five cards from a shuffled deck, how many
possible hands can be dealt? The area of mathematics that deals with counting
techniques is called combinatorics.
The original purpose for the development of counting techniques was to determine
the appropriate odds, payoffs, and expected earnings for games of chance. As applied
to casinos and lotteries, this use of counting is still one of the most common applications
of combinatorics. We shall begin our study of counting with two fundamental
principals.
The Principal of Choice
Let A be a set with N distinct objects. Then there are N possible
ways to choose exactly one object from set A .
If we allow ourselves the additional choice of not choosing an
object, then there are N + 1 possible choices.
If we choose one object at random from set A so that each object is
equally likely to be chosen, then the probability of choosing any
single object is 1 / N .
Example 1. Suppose you have 41 pairs of shoes and you are trying to decide what to
wear.
(a) In theory, how many choices do you have?
(b) Are all of these choices really equally likely? If you do grab a pair of shoes at
random, then what is the probability of grabbing your favorite pair?
(c) In reality, do you actually have the same number of choices if you are going to the
swimming pool as compared to going out on a formal dinner date?
Solution. (a) In theory, you have 42 choices because you might decide to go barefoot.
(b) Depending on where you are going, the choices probably are not equally likely. If
you were to choose a pair at random though, then each pair of shoes would have a 1/41
chance of being picked.
(c) If you are going to the pool, then your choices most likely are narrowed down to
barefoot, sandals, flip-flops, or sneakers. You normally would choose from a different
set of options (dress shoes) for a formal dinner date.
Example 2. A vending machine has 20 different types of snacks, including Ruffles. You
are thinking about buying a snack.
(a) How many choices do you have?
(b) If you do decide to buy a snack and you choose one at random, then what is the
probability of choosing Ruffles? Explain what this probability really means.
Solution. (a) You actually have 21 choices because you may decide not to buy anything.
(b) If you do choose one at random, then there is a 1/20 chance. of picking Ruffles. That
is, if you were to pick one of these snacks at random, over and over, many times, then
about 5% of the time you would pick Ruffles.
The Multiplication Principal
1. Let A be a set with N distinct objects. If we repeatedly choose objects
from set A with repeats allowed, a total of k times, then there are
N ! N !. .. ! N = N
k
possible outcomes for the ordered sequence of k choices.
2. Let A1 be a set with N1 distinct objects, let A2 be a set with N2 distinct
objects, . . . , let Ak be a set with Nk distinct objects. If we choose one
object from each of the sets A1 , A2 , . . ., Ak , then there are
N1 ! N2 ! . . . ! N k
possible outcomes for the sequence of choices.
Example 3. How many phone numbers are there of the form 745-xxxx?
Solution. For each of the last four digits, there are 10 choices (from 0 – 9). The digits are
4
ordered with repeats allowed; thus, there are 10 ! 10 ! 10 ! 10 = 10 = 10,000 such phone
numbers.
Example 4. You are choosing an 8-character alpha-numeric password. You must
choose characters from a–z, 0–9, or from 11 other symbols such as !, @, #, etc. The code
is not case-sensitive and you may repeat characters.
(a) How many possible passwords are there?
(b) How many passwords have 5 digits (from 0–9) followed by 3 letters?
Solution. (a) For each character you may choose from 26 letters, 10 digits, and 11
symbols giving 47 choices. Since there are 8 characters which are ordered with possible
8
13
repeats, there are a total of 47 ! 47 ! 47 ! 47 ! 47 ! 47 ! 47 ! 47 = 47 ≈ 2.3811287 ! 10
such passwords.
5
3
(b) There are 10 ! 26 = 1,757,600,000 passwords with 5 digits followed by 3 letters.
Example 5. In the 2012 election, the ballot in one precinct will have races for
President/Vice President (4 candidates), U.S. Senate (3 candidates), U.S. House (2
candidates), District Judge (4 candidates), and Mayor (5 candidates).
Assuming you do not vote for any write-in candidates, how many possible choices
do you have for casting this ballot?
Solution. Assuming that we actually vote for someone in each race, then we have
4 ! 3 ! 2 ! 4 ! 5 = 480 choices. By allowing ourselves the extra choices of not voting in
some (or all) races, then we have 5 ! 4 ! 3 ! 5 ! 6 = 1800 choices.
Choosing an Ordered Sequence from a Set
With No Repeats
Let A be a set with N distinct objects. If we choose a sequence of objects from set A in
order with no repeats, then on each successive choice we have one less choice.
Example 6. A Master lock uses three numbers from 0 – 39 without repeats. How many
possibilities are there?
Solution. Including the 0, there are 40 choices for the first digit. These choices decrease
by one for each successive digit because repeats are not allowed. Also the sequence of
numbers is ordered; thus, there are 40 ! 39 ! 38 = 59,280 possible combinations for the
lock.
Example 7. A state’s license plate is of the form CBE 925, (three letters followed by three
digits). (a) How many such plates with this form are possible? (b) What percentage of
these plates have no letter or digit repeated?
Solution. (a) In theory, all letters and digits may be repeated; so there are
3
3
26 ! 26 ! 26 ! 10 ! 10 ! 10 = 26 ! 10 = 17,576,000 possible license plates of this form.
(b)
The number of plates with no repeated letters or digits is given by
26 ! 25 ! 24 ! 10 ! 9 ! 8 = 11,232,000 for a proportion of 11,232,000/17,576,000 ≈ 0.639.
So around 63.9% of these license plates do not have any repeated letters or digits. (And
around 36.1% will have a repeated letter or digit.)
Example 8. What percentage social security numbers have a repeated digit?
Solution. There are nine digits in a social security number; so theoretically there are 10
(one-billion) possible SSNs. The number of cards without a repeated digit is
9
10 ! 9 ! 8 ! 7 ! 6 ! 5 ! 4 ! 3 ! 2 = 3,628,800.
So there are 10 9 ! 3, 628, 800 = 996,371,200 cards with a repeated digit. Thus the
996,371, 200
proportion of cards with a repeated digit is
= 0.9963712. That is, about
109
99.637% of social security numbers have a repeated digit.
Exercises
1. Suppose you have 40 HD channels in your cable package including ESPN.
(a) When you plop down in front of the set, how many choices do you have in choosing
an HD channel to watch?
(b) If you choose an HD channel at random, then what is the probability of ESPN being
chosen? Explain what this probability means.
2. There are 52 cards in a standard deck. If you choose one card at random, then how
many choices do you have? What is the probability of each choice?
3. In the SGA elections, there are 3 people running for President and 4 people running
for Treasurer. How many choices do you having for casting a ballot for these two
offices?
4. To complete the General Education requirements, a student needs one course in each
of Foreign Language (Cat. A. II), Mathematics (Cat. D. II) and Physical Development
(Cat. E.). There are 13 foreign language courses, 7 mathematics courses, and 10 physical
development courses. In theory, how many choices does the student have in choosing
three courses to satisfy these Gen. Ed. requirements?
5. A combination lock has 4 tumblers each with the numbers 0 – 14.
(a) How many possible combinations for the lock are there if numbers can be repeated?
(b) How many of these combinations do not repeat any numbers?
(c) What percentage of the combinations have repeated numbers?
6. A serial number has 4 digits (chosen from 0–9), followed by 3 consonants, followed
by 2 vowels.
(a) How many such serial numbers are possible if repeated characters are allowed?
(b) What percentage of these serial numbers have no repeated characters?
7. In a lottery game, players must match five numbers chosen from 1 – 40, and then
match the Lucky Powerball in the sixth spot chosen from 1 – 30. How many choices are
there for the six numbers, if in the first five places
(a) numbers are in order with repeats allowed?
(b) numbers must be in order with no repeats?
Solutions
To receive maximum partial credit on the exam in the case of an incorrect answer, it is
best to show the terms that you are multiplying and then give the final answer. For
example, 10 ! 9 ! 8 = 720.
1. (a) If you include the option of not choosing an HD channel, then you have 41
choices.
(b) But if you do choose an HD channel at random, then each channel, including ESPN,
has probability 1/40 of being chosen. That is, if you were to choose an HD channel at
random, over and over, many times, then about 2.5% of the time you would choose
ESPN.
2. When choosing a card at random, there are 52 possible outcomes each occurring with
probability 1/52.
3. If you actually cast a vote for each office, then you have 3 " 4 = 12 choices. But in
reality, you have the extra choices of not voting for one or both offices which actually
gives you 4 " 5 = 20 choices.
!
4. By choosing one course from each category, the student has 13 ! 7 ! 10 = 910 ways of
!
finishing
up these Gen. Ed. requirements.
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5. (a) If numbers can be repeated, then there are 15 ! 15 ! 15 ! 15 = 15 = 50,625 possible
combinations.
(b) The number of combinations without repeated numbers is 15 ! 14 ! 13 ! 12 = 32,760.
(Thus there are 50,625 – 32,760 = 17,865 combinations that have a repeated number.)
(c) 17,865/50,625 ≈ 0.352888 . . . So around 35.29% of the possible combinations have a
repeated number.
6. (a) With repeats allowed, there are (10 ! 10 ! 10 ! 10) ! (21! 21 ! 21) ! (5 ! 5) =
2,315,250,000 possible serial numbers.
(b) There are (10 ! 9 ! 8 ! 7) ! (21 ! 20 ! 19) ! (5 ! 4) = 804,384,000 of these serial numbers
having no repeated characters. Thus around 34.74% of these serial numbers have no
repeated characters.
7. (a) If in the first five places the numbers are in order with repeats allowed, then
5
there are 40 ! 30 = 3,072,000,000 choices. (over 3 billion)
(b) If there are no repeats allowed, then there are (40 ! 39 ! 38 ! 37 ! 36) ! 30 =
2,368,828,800 choices. (nearly 2.4 billion)