Rule of Sum
If a task can be done either in one of n1 ways or
in one of n2 ways, where none of the set of n1
ways is the same as any of the set of n2 ways,
then there are n1 + n2 ways to do the task
Suppose that either a member of the mathematics
faculty or a student who is a mathematics major
is chosen as a representative to a university
committee. How many different choices are there
for this representative if there are 37 members
of the mathematics faculty and 83 mathematics
majors and no one is both a faculty member and a
student?
120 = 83 + 37
A student can choose a computer project from one
of three lists. The three lists contain 23, 15,
and 19 possible projects, respectively. No
project is on more than one list. How many
possible projects are there to choose from?
57 = 23 + 15 + 19
Rule of Product
Suppose that a procedure can be broken down into
a sequence of two tasks. If there are n1 ways to
do the first task and for each of these ways of
doing the first task, there are n2 ways to do the
second task, then there are n1n2 ways to do the
procedure.
An interesting problem with an interesting
solution: how numbers divide evenly the number
957?
957 = 3 x
2 numbers
2 numbers
2 numbers
319 = 3 x 11 x 29
divide 3 – 3 and 1
divide 11 – 11 and 1
divide 29 – 29 and 1
8 = 2 x 2 x 2
1,3,11,29,33,87,319,957
How many numbers divide 57?
57 = 3 x 19
2 numbers divide 3
2 numbers divide 19
4 numbers divide 57
1,3,19,57
How many numbers divide 100?
100 2 x 2 x 5 x 5 = 22 x 52
3 numbers divide 4 – 1,2 and 4
3 numbers divide 5 – 1,5 and 25
9 = 3 x 3 numbers divide 100
1,2,4,5,10,20,25,50,100
A new company with just two employees, Sanchez
and Patel, rents a floor of a building with 12
offices. How many ways are there to assign
different offices to these two employees?
Suppose there are only 3 offices:
p1s2, p1s3, p2s1, p2s3, p3s1, p3s2
There are 132 = 12 x 11 ways to assign different
offices to Patel and Sanchez
The chairs of an auditorium are to be labeled
with a letter and a positive integer not
exceeding 100. What is the largest number of
chairs that can be labeled differently?
There are 26 ways to choose the letter
There are 100 ways to choose the number
Are there 2600 hundred ways to choose the letter
and the number? What does label look like. Is
this an example of a label A99 or is this an
example of a label 99A?
How many labels are there if the labels are
letter or digits but not both letter and digits?
126 = 100 + 26
There are 32 microcomputers in a computer
center. Each microcomputer has 24 ports. How
many different ports to a microcomputer in the
center are there?
C1P1
…
C1P24
C2P1
…
C2P24
…
C32P24
32 x 24 = 32 x 8 x 3 = 768
How many different bit strings of length seven
are there?
How many different license plates are available
if each plate contains a sequence of three
letters followed by three digits (and no
sequences of letters are prohibited)?
{A,B}, (1,2}
2 x 2 x 2 x 2
AA11,
AA12,
AA21,
AA22,
BB11,
BB12,
BB21,
BB22,
AB11,
AB12,
AB21,
AB22,
BA11
BA12
BA21
BA22
26 x 26 x 26 x 10 x 10 x 10
Rule of Product and Rule of Sum
Each user on a computer system has a password,
which is six to eight characters long, where
each character is an uppercase letter or a
digit. Each password must contain at least one
digit. How many possible passwords are there?
P6, P7, P8
# of passwords = |P6| + |P7| + |P8|
There 36 x 36 x 36 x 36 x 36 x 36 = 366
unrestricted pass words. How many passwords
letters and no digits?
266
What does 366 – 266 represent? It is the number
of passwords that have at least one digit!
|P6| = 366 - 266
|P7| = 367 - 267
|P8| = 368 - 268
How many license plates can be made using either
three digits followed by three letters or three
letters followed by three digits?
3L3D
3D3L
|3L3D| + |3D3L| = ?
|3L3D| = 26 x 26 x 26 x 10 x 10 x 10
|3D3L| = 26 x 26 x 26 x 10 x 10 x 10
# of license plates = |3L3D| + |3D3L|
Permutations
In how many ways can we select five students to
stand in line for a picture?
{A,B} AB, BA
{A,B,C} ABC, ACB, BAC, BCA, CAB, CBA
120 = 5 x 4 x 3 x 2 x 1 ways for the students to
stand in line
In how many ways can we select three students
from a group of five students to stand in line
for a picture?
{A,B,C,D,E}
60 = 5 x 4 x 3 ways to select 3 students from 5
students to stand in line
How many ways are there to select a first-prize
winner, a second-prize winner, and a third-prize
winner from 100 different people who have
entered a contest?
100 x 99 x 98
Suppose that there are eight runners in a race.
The winner receives a gold medal, the secondplace finisher receives a silver medal, and the
third-place finisher receives a bronze medal.
How many different ways are there to award these
medals, if all possible outcomes of the race can
occur and there are no ties?
336 = 8 x 7 x 6
Suppose that a saleswoman has to visit eight
different cities. She must begin her trip in a
specified city, but she can visit the other
seven cities in any order she wishes. How many
possible orders can the saleswoman use when
visiting these cities?
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
How many permutations of the letters ABCDEFGH
contain the string ABC?
Let X be ABC then there are only 6 permutable
symbols
X,D,E,F,G,H
6! = 6 x 5 x 4 x 3 x 2 x 1
P(n,r) = n!/(n – r)!
P(n,r) = n x n – 1 x … x (n – r + 1)
0, 1, … r – 1 represent r factors
# of factors = n – (n – r + 1) + 1
= n – n + r – 1 + 1 = r
P(5,3) = 5!/(5 – 3)! = 5!/2! =
5 x 4 x 3 x 2 x 1 / (2 x 1) = 60
P(5,3) = 5 x 4 x 3 = 60
Combinations
P(n,r) = n!/(n – r)!
{A, B, C}
ABC,ACB,BAC,BCA,CAB,CBA
or 6 permutations
C(4,3) = ?
A,B,C,D
ABC -> {ABC, ACB, BAC, BCA, CAB, CBA}
ABD -> {ABD, ADB, BAD, BDA, DAB, DBA}
BCD -> {BCD, BDC, CBD, CDB, DBC, DCB}
ACD -> {ACD, ADC, CAD, CDA, DAC, DCA}
C(4,3) x 3! = P(4,3)
C(4,3) = P(4,3)/3!
C(n,3) = P(n,3)/3!
C(n,r) = P(n,r)/r!
r! = r x (r – 1) x ... 2 x 1
How many different committees of three students
can be formed from a group of four students?
C(4,3) = P(4,3)/3! = 4!/1!3! =
4 x 3 x 2 x 1 /(1 x 3 x 2 x 1) = 4
ABC,ABD,ACD,BCD
How many poker hands of five cards can be dealt
from a standard deck of 52 cards? Also, how many
ways are there to select 47 cards from a
standard deck of 52 cards?
C(52, 5)
C(52,47)
C(52,5) = C(52,47)?
52!/((52-5)!5!) = 52!/(47!5!)
52!/((52-47)!47!) = 52!/(5!47!)
C(n,r) = C(n, n - r)
How many ways are there to select five players
from a 10-member tennis to make a trip to a
match at another school?
What is n? n is 10
What is r? r is 5
C(10,5) = 10!/(5!5!)
= 10 x 9 x 8 x 7 x 6
= 10 x 9 x 8 x 7 x 6
= 9 x 8 x 7 x 6 / (4
= 3 x 8 x 7 x 6 / (4
= 3 x 2 x 7 x 6
= 252
/(5 x 4 x 3 x 2 x 1)
/(5 x 4 x 3 x 2 x 1)
x 3 x 1)
x 1)
A group of 30 people have been trained as
astronauts to go on the first mission to Mars.
How many ways are there to select a crew of six
people to go on this mission (assuming that all
crew members have the same job)?
C(30, 6)
Suppose that there are 9 faculty members in the
mathematics department and 11 in the computer
science department. How many ways are there to
select a committee to develop a discrete
mathematics course at a school if the committee
is to consist of three faculty members from the
mathematics department and four from the
computer science department?
C(9,3) ways to select the mathematics members
C(11,4) ways to select the comp science members
# of committees is C(9,3) x C(11,4)
C(11,4) = 11!/((11 – 4)!4!)
C(n,r) = P(n,r)/r!
= n!/(n-r)!/r!
= n!/r!(n-r)!
A coin is flipped eight times where each flip
comes up either heads or tails. How many
possible outcomes are there in total? How many
outcomes contain exactly two heads? How many
outcomes contain exactly three tails? How many
outcomes contain at least three heads? How many
outcomes contain the same number of heads as
tails?
256 = 28 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2
C(8,2) outcomes contain exactly 2 heads
C(8,3) outcomes contain exactly 3 tails
|at least 3 heads| =
C(8,8)+C(8,7)+C(8,6)+C(8,5)+C(8,4)+ C(8,3)
|same number of heads as tails| = C(8,4)
Simplified versions
C(n,r) = n!/((n-r)!r!)
n! = n x (n-1) x (n-2) x... 2 x 1
C(4,2) =
=
=
=
4!/((4 – 2)!2!) = 4!/(2!2!)
4 x 3 x 2 x 1/(2 x 1 x 2 x 1)
4 x 3 x 2 x 1/4
6
HHHH,
HHHT,
HHTH,
HHTT,
C(4,3)
HHHH,
HHHT,
HHTH,
HHTT,
HTHH,
HTHT,
HTTH,
HTTT,
THHH,
THHT,
THTH,
THTT,
TTHH
TTHT
TTTH
TTTT
= 4!/((4-3)!3!) = 4!/(1!3!)
= 4!/3!
= 4 x 3 x 2 x 1 /(3 x 2 x 1)
= 4
HTHH,
HTHT,
HTTH,
HTTT,
THHH,
THHT,
THTH,
THTT,
TTHH
TTHT
TTTH
TTTT
If we flip 4 times how many ways can we get the
same number of heads as tails?
C(4,2) = 6
HHHH,
HHHT,
HHTH,
HHTT,
HTHH,
HTHT,
HTTH,
HTTT,
THHH,
THHT,
THTH,
THTT,
TTHH
TTHT
TTTH
TTTT
Seven women and nine men are on the faculty in
the mathematics department at a school. How many
ways are three to select a committee of five
members of the department If at least one woman must be on the committee?
C(16,5) – C(7,5)
How many ways are there to select a committee of
five members if the department if at least one
woman and at least one man must be on the
committee?
C(16,5) – C(9,5) – C(7,5)
Permutations and Combinations
Thirteen people on a softball team show up for a
game. How many ways are there to choose 10
players to take the field?
C(13,10)
How many ways are there to assign the 10
positions by selecting players from the 13
people who show up?
C(13,10)10! = P(13,10)