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Lecture 2.5
Contemporary Mathematics
Instruction: Linear versus Circular Permutations
Consider eight people who must form a line. How many ways can the people be arranged
in a line? There are eight ways to determine the first position. After one person occupies the first
position, there are only seven people who can occupy the second position. After two positions
are occupied, there are only six people who can occupy the third position, etc. Using the
Fundamental Counting Principle, the total number of ways the eight people can form a line is:
8·7·6·5·4·3·2·1 = 8! = 40,320. This example demonstrates the Factorial Rule, which states that
the number of permutations when n objects are chosen from a set of n objects equals n!
The example above is a linear arrangement where objects are assigned absolute positions.
In circular arrangements, the objects are arranged relative to one another and the first object
placed provides a point of reference. Consider set A = {,,e}. There are six linear
arrangements of three objects taken from set A as shown with the permutations below.
,,e
,e,
,,e
,e,
e,,
e,,
In contrast to linear arrangements, there are only two circular arrangements of three objects taken
from set A as shown with the circular permutations shown in Figure 1 and 2 below.
e Figure 1
e
Figure 2
At first glance a student might wonder why Figure 3 below should not also be counted as a
different arrangement, but objects have relative positions in circular permutations, and the
relative positions in Figure 3 are no different than the relative positions in Figure 2.
e Figure 3
To find the number of circular arrangements of all the objects in a set of n distinct objects,
choose one object to occupy a fixed position, the remaining n − 1 objects will be arranged using
this fixed position as reference in a ring, so there are ( n − 1) ! ways of arranging these n − 1
objects by the Factorial Rule.
For example, how many ways can eight people be arranged in a ring. In a set of eight
people, n = 8 and n – 1 = 7, so the number of circular permutations of eight people is 7!
7 × 6 × 5 × 4 × 3 × 2 × 1 = 5, 040
Lecture 2.5
The above examples are special cases of circular permutations of r objects taken from a
set of n objects where r = n. In general, the number of circular permutations is given by the
Circular Permutation Formula below.
If set S has n elements, then the number of circular arrangements containing r
distinct objects taken from S is denoted by CP(n,r) or nCPr and given by
. CP ( n, r ) =
n!
r (n − r )!
In particular, the Circular Permutation Formula gives the number of ways to arrange four of the
eight people in a ring:
2
8!
8 × 7 × 6 × 5 × 4!
CP ( 8, 4 ) =
=
= 2 × 7 × 6 × 5 = 420
4 (8 − 4) !
4 ( 4) !
Application Exercise 2.5
Problems
#1
Compute CP (15, 6 ) .
#2
An electrical engineer has six available resistors that meet the specifications for a
particular circuit board where four resistors are placed relative to one another. In how
many ways can the engineer arrange four of these resistors in the circuit?
#3
An engineer is designing a merry-go-round. The construction of the carousel makes
room for ten platforms (or seats). The engineer plans to use wooden figures taken from
old, moth-balled carousels for the seats. The engineer has fourteen such figures. In how
many ways can ten of these fourteen figures be arranged on the new merry-go-round?
1. 600,600
2. 90
3. 363,242,880
Assignment 2.5
Problems
#1
Compute CP (12, 7 ) .
#2
How many ways can six dinner guests be seated relative to one another at a round
table?
#3
Emily has four dolls that she likes to arrange in a line. Constance has five dolls but she
prefers a circular arrangement. Which girl has more arrangements for her dolls?
#4
There are seven guests for dinner. The hostess seats six at a circular table, forcing one
guest to sit at an adjoining card table. In how many ways can the guests be positioned
relative to one another.
#5
A baseball "diamond" looks like the diagram below.
If we have four player and want to count the number of arrangements for putting 1
player at 1st base, 1 player at 2nd base, 1 player at 3rd base, and 1 player at home,
should we use P ( 4, 4 ) or CP ( 4, 4 ) ? Explain.
Lecture 2.6
Contemporary Mathematics
Instruction: Special Permutations with Repetition
Readers may recall the special warning from section 2.3 that cautioned, "Since the
Permutation Formula gives the number of arrangements of r distinct objects taken from a set of
n distinct objects, there is no duplication of objects." This section, however, deals with special
permutations where the repetition of objects may occur.
A moment must be taken to discuss repetition. The phrase "repetition may occur," means
that some objects may appear the in the arrangement more than once. For our purposes,
however, we will restrain repetition to a strict definition. Consider set A = {d , o, z} and the
arrangement of the elements in set A that forms the word "zoo." In this arrangement, the letter z
appears once, the letter o appears twice, and the letter d does not appear. Each appearance is said
to be a repetition by the definition stated below so that in the word "zoo," z is repeated once, o is
repeated twice, and d is not repeated.
A repetition is the appearance of an object from a set in an arrangement.
Using this definition, each appearance of an object in an arrangement is called a repetition so
that an object that does not appear in the arrangement is said to be repeated zero times. An object
that appears only once in the arrangement is said to be repeated once. An object that appears
twice in the arrangement is said to have been repeated twice, and so on.
Assume S is a set of n distinct objects. Consider the arrangements where r1 repetitions are
required for the first object in S, r2 repetitions are required for the second object in S, and so on
to rn repetitions required for the n-th object in S, so that r1 + r 2 + " + rn = rt , and rt is the total
number of repetitions, which by definition is equivalent to saying rt is the total number of
objects (not necessarily distinct) in the arrangement. The number of these possible arrangements
is given by the Permutation of Non-Distinct Objects Formula given below.
rt !
= the number of arrangements with rt objects not necessarily distinct
r1 ! r2 !" rn !
taken from a set of n objects where the first element appears r1 times, the second
element appears r2 times, . . ., and the nth element is repeated rn times.
Turning to an example, consider the set of digits in a binary number system, U = {0, 1} .
Consider the number of possible arrangements in which each digit must be repeated 4 times.
Since the two objects in U will each be repeated four times, any one arrangement in the set of
Lecture 2.6
arrangements will have eight objects, so r = 8. The number of arrangements (permutations)
possible is calculated below.
2
8!
8 × 7 × 6 × 5 × 4! 8 × 7 × 6 × 5 8 × 7 × 5
=
= 2 × 7 × 5 = 70
=
=
4!4!
4 ×1
4! × 4 × 3 × 2 × 1 4 × 3 × 2 ×1
rt !
reveals
r1 ! r2 !" rn !
that any of the factorials in the denominator of the formula that represent non-usage--that is, zero
repetition--of an element can be ignored since 0! = 1. Similarly, any of the factorials in the
denominator of the formula that represent a single use--that is, a single repetition--of an element
can be ignored since 1! = 1. The following example involving anagrams applies this observation
directly.
Examination of the Permutation of Non-Distinct Objects Formula
An anagram is a word or phrase formed by transposing the letters of another word or
phrase. The phrase "Twelve add one" is an anagram of "Eleven add two." Consider the task of
searching for other anagrams of "Eleven add two." An anagram is formed by transposing or rearranging the letters in the given phrase. The task of finding the anagrams presents a counting
question, "How many different arrangements of the letters would have to be examined to find all
the anagrams of the phrase?" To answer this question, consider the letters in the phrase "Eleven
add two" as a set, E = {a, d , e, l , n, o, t , v, w} . Anagrams of "Eleven add two" are arrangements
of the elements in E that require three repetitions of e, two repetitions of d, and only one
repetition of each of the remaining elements. The number of permutations to examine can be
counted using the Permutation of Non-Distinct Objects Formula as shown below.
2
12!
12 ×11× 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3!
=
= 12 × 11× 10 × 9 × 8 × 7 × 6 × 5 × 2 = 39, 916, 800
3!2!
3! × 2 × 1
In each of the two previous examples, a determined number of repetitions was stated. A
natural question that arises asks, "What if an arrangement had a set number of objects but
unlimited repetitions?" The Fundamental Counting Principle addresses this question. Consider
a set of n distinct objects that must be arranged into arrangements of p objects. Think of placing
an object in each position of the arrangement as a task. There are n ways to fill the first position
(because all n objects can be placed in the first position), n ways to fill the second position
(because all n objects can still be placed in the second position since repetition is allowed and
unlimited), n ways to fill the third position, and so on until the p-th position. Applying the
Fundamental Counting Principle, the number of arrangements equals n p .
For an example, consider the number of four digit pass-codes that can be formed. The
pass-codes are composed of digits, so n = 10. Since the pass-codes are comprised of only four
digits, the number of pass-codes possible equals 104 = 10,000 .
Application Exercise 2.6
Problems
#1
Morse code employs short and long dashes. Assuming that any arrangement of short
and long dashes could convey a message, how many messages are possible using three
short dashes and four long dashes?
#2
Binary numbers are formed using only two digits, zero and one. How many numbers
can be formed using an arrangement of five binary digits with two zeros and three
ones?
#3
If NASA plans nine shuttle missions using Atlantis, Discovery, and Endeavour each
three times, in how many ways can the nine missions be arranged?
#4
Consider set B such that n ( B ) = 6 . How many permutations of the objects in B are
possible if each object is repeated twice?
#5
Consider set R such that n ( R ) = 4 . How many six-object permutations of the elements
in R are possible if repetition is allowed and unlimited?
1. 35
2. 10
3. 1,680
4. 7,484,400
5. 4,096
Assignment 2.6
Problems
#1
Consider set B = {←, →, ↔} . How many permutations are possible using two left
arrows, two right arrows, and three double arrows?
#2
Morse code employs short and long dashes. Assuming that any arrangement of short
and long dashes could convey a message, how many messages are possible using four
short dashes and five long dashes?
#3
Binary numbers are formed using only two digits, zero and one. How many numbers
can be formed using an arrangement of four binary digits with one zero and three ones?
#4
Consider set G such that n ( G ) = 5 . How many permutations of the objects in G are
possible if each object is repeated twice?
#5
Consider set H such that n ( H ) = 5 . How many six-object permutations of the elements
in H are possible if repetition is allowed and unlimited?