Objective
Students will use the rules for counting.
Fundamental Counting Rule
For a sequence of two events in which the
first event can occur m ways and the
second event can occur n ways, the events
together can occur a total of mn ways.
This can be generalized to any finite
number.
Example
Computers are designed so that the most basic unit of
information is a bit (or binary digit) which represents
either 0 or 1. Letters, digits, and punctuation symbols are
represented as bytes. A byte is a sequence of eight bits in
a particular order. For example The ASCII coding system
uses 01000001 to represent the letter A and 00110111 to
represent the digit 7. How many characters can be
represented using a byte system?
(How many possible sequences of 8 are there?)
Solution
• This is a FCR problem.
• ________
• There are 2 choices for the first, 2 choices
for the second, 2 choices for the third, and
so on up to the eighth bit. There are 256
possible bytes for 256 different characters.
2  256
8
Permutations and Combinations
• Does Order Matter?
• A permutation of objects is a sequence of
objects where order matters.
• A combination of objects is a sequence of
objects where order doesn’t matter.
• Think permutation position and combination
committee.
Permutations and Combinations
• abc and bca are the same combination.
• abc and bca are not the same permutation.
• Are there more permutations of the elements
{a,b,c} or combinations of them and why?
• There are more permutations because each
triple is unique.
Example
• Find all possible combinations of 2 elements of
{a,b,c}.
• Find all possible permutations of 2 elements of
{a,b,c}.
Solution
• The possible combinations of 2 elements of
{a,b,c} are {a,b} {a,c} and {b,c}. 3 combinations
exist.
• The possible permutations of 2 elements of
{a,b,c} are (a,b) ,(a,c) ,(b,a) ,(c,a), (b,c) and,
(c,b). 6 permutations exist.
Definition
The factorial symbol is !
n! means n(n-1)(n-2)…(3)(2)(1). It is a product of
decreasing positive whole numbers.
For example,
4!  4  3  2 1  24
By special definition, 0! = 1.
Permutations Rule
If there are n different items available and we select r of the n items (without
replacement), the number of permutations (or sequences) of r items selected from n
available items (without replacement) is
n!
n Pr 
(n  r )!
NOTE
1.We use a different rule if some of the items are identical to others.
2. We consider rearrangements of the same items to be different sequences.
(The permutation of ABC is different from CBA and is counted separately.)
Combinations Rule
If there are n different items available and we select r of the n items (without
replacement), the number of combinations (or sequences) of r items
selected from n available items (without replacement) is
n!
n Cr 
(n  r )!r !
We are dividing by an extra r!
NOTE
•
We consider rearrangements of the same items to be the same. (The
combination of ABC is the same as CBA.)
Example
• Find all possible
combinations of 2
elements of {a,b,c}
using the formula.
n!
n Cr 
(n  r )!r !
• Find all possible
permutations of 2
elements of {a,b,c}
using the formula.
n!
n Pr 
(n  r )!
Factorial Rule
(The FCR in disguise)
For a sequence of n events in which the first
event can occur n ways and the second
event can occur n-1 ways, the third event
can occur n-2, the events together can
occur a total of n! ways.
So the number of different permutations of n
different items is n!
Example
A history pop quiz has a question which asks students to
arrange the following presidents in chronological order:
Hayes, Taft, Polk, Taylor, Grant, Pierce.
If an unprepared student totally guesses, what is the
probability of guessing correctly?
Possible arrangements: 6!  720
1
P  guessing correctly  
 0.00139
720
Solution
• This is an FCR problem (without
replacement).
• ______
• There are 6 choices for the first, 5 choices
for the second, 4 choices for the third, and
so on all the way to the 6th choice.
• 6x5x4x3x2x1 = 720 is the sample space
and there is only 1 correct answer. 1/720 =
0.00139. The probability of a correct
answer is 0.00139. This student is in
trouble…
Example
• In horse racing a bet on an exacta in a race is
won by correctly selecting the horses that finish
first and second, and you must select the horses
in correct order. The 136th running of the
Kentucky derby had a field of 20 horses. If a
bettor randomly selects two of those horses for
an exacta bet, what is the probability of winning
by selecting Super Saver and Ice Box as the first
and second? Do all exacta bets have the same
probability of winning?
Solution
• This is a permutations problem of 20P2.
We must determine the number of
possible permutations of 2 items taken 20.
• Our formula is
n!
n Pr 
(n  r )!
where n is 20 and r is 2.
Solution continued…
• 20!/18! = 380. Our sample space consists
of 380 possible outcomes. Since only 1
pair of horses is Super Saver, Ice Box the
probability of selecting that winning
arrangement is 1/380.
• Not all pairings of horses have the same
probability of winning because some are
faster than others.
The Factorial Rule
(with repeated elements)
If repetitions occur, we must divide by the
factorials of the quantities of the repetitions.
If there are n1 alike, n2 alike, . . . nk alike, the
number of permutations (or sequences) of all items
selected without replacement is
n!
n1 !n2 ! nk !
This is the number of different permutations of n
items with k repetitions.
Example
•
•
•
•
For the set of numbers find 6C3 and 6P3.
{1,2,3,4,5,6}
6C3 = 6!/(6-3)!3! =6!/3!3!=720/6x6=20
6P3 = 6!/(6-3)! = 720/6=120
Example
• Find the number of permutations of the
set.
• {1,1,1,2,3,4}
• 6!/3! = 720/6 = 120
Example
In the Pennsylvania Match 6 Lotto, winning the jackpot
requires you select six different numbers from 1 to 49.
The winning numbers may be drawn in any order. Find
the probability of winning if one ticket is purchased.
n!
49!
Number of combinations: nCr 

 13,983,816
 n  r !r ! 43!6!
1
P  winning  
13,983,816
Quiz
• Find all combinations of 2 elements of {a,b,c,d}
• Find all permutations of 2 elements of {a,b,c,d}
• A thief steals an ATM card and must randomly guess
the pin (of four digits out of a possible 0-9). What is
the probability of the thief guessing the pin on one try?