Math 1300: Section 7- 4 Permutations and Combinations

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FactorialsPermutationsCombinations

Math 1300 Finite MathematicsSection 7-4: Permutations and Combinations

Jason Aubrey

Department of MathematicsUniversity of Missouri

Jason Aubrey Math 1300 Finite Mathematics

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Problem 1: Consider the set {p,e,n}. How many two-letter"words" (including nonsense words) can be formed from themembers of this set, if two different letters have to be used?

We list all possibilities: pe, pn, en, ep, np, ne, a total of 6.

Problem 2: Now consider the set consisting of three males:{Paul ,Ed ,Nick}. For simplicity, denote the set by {p,e,n}.How many two-man crews can be selected from this set?

pe (Paul, Ed), pn (Paul, Nick) and en (Ed, Nick)


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Both problems involved counting the numbers of arrangementsof the same set {p,e,n}, taken 2 elements at a time, withoutallowing repetition.

However, in the first problem, the order ofthe arrangements mattered since pe and ep are two different"words". In the second problem, the order did not matter sincepe and ep represented the same two-man crew. We countedthis only once. The first example was concerned with countingthe number of permutations of 3 objects taken 2 at a time. Thesecond example was concerned with the number ofcombinations of 3 objects taken 2 at a time.


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Both problems involved counting the numbers of arrangementsof the same set {p,e,n}, taken 2 elements at a time, withoutallowing repetition. However, in the first problem, the order ofthe arrangements mattered since pe and ep are two different"words".

In the second problem, the order did not matter sincepe and ep represented the same two-man crew. We countedthis only once. The first example was concerned with countingthe number of permutations of 3 objects taken 2 at a time. Thesecond example was concerned with the number ofcombinations of 3 objects taken 2 at a time.


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Both problems involved counting the numbers of arrangementsof the same set {p,e,n}, taken 2 elements at a time, withoutallowing repetition. However, in the first problem, the order ofthe arrangements mattered since pe and ep are two different"words". In the second problem, the order did not matter sincepe and ep represented the same two-man crew. We countedthis only once.

The first example was concerned with countingthe number of permutations of 3 objects taken 2 at a time. Thesecond example was concerned with the number ofcombinations of 3 objects taken 2 at a time.


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Both problems involved counting the numbers of arrangementsof the same set {p,e,n}, taken 2 elements at a time, withoutallowing repetition. However, in the first problem, the order ofthe arrangements mattered since pe and ep are two different"words". In the second problem, the order did not matter sincepe and ep represented the same two-man crew. We countedthis only once. The first example was concerned with countingthe number of permutations of 3 objects taken 2 at a time.

Thesecond example was concerned with the number ofcombinations of 3 objects taken 2 at a time.


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Both problems involved counting the numbers of arrangementsof the same set {p,e,n}, taken 2 elements at a time, withoutallowing repetition. However, in the first problem, the order ofthe arrangements mattered since pe and ep are two different"words". In the second problem, the order did not matter sincepe and ep represented the same two-man crew. We countedthis only once. The first example was concerned with countingthe number of permutations of 3 objects taken 2 at a time. Thesecond example was concerned with the number ofcombinations of 3 objects taken 2 at a time.


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Definition (Factorial)For n a natural number,

n! = n(n − 1)(n − 2) · · · 2 · 10! = 1n! = n · (n − 1)!

Note: Many calculators have an n! key or its equivalent


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Examples3! = 3(2)(1) = 6

4! = 4(3)(2)(1) = 125! = 5(4)(3)(2)(1) = 1206! = 6(5)(4)(3)(2)(1) = 7207! = 7(6)(5)(4)(3)(2)(1) = 5040


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Examples3! = 3(2)(1) = 64! = 4(3)(2)(1) = 12

5! = 5(4)(3)(2)(1) = 1206! = 6(5)(4)(3)(2)(1) = 7207! = 7(6)(5)(4)(3)(2)(1) = 5040


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Examples3! = 3(2)(1) = 64! = 4(3)(2)(1) = 125! = 5(4)(3)(2)(1) = 120

6! = 6(5)(4)(3)(2)(1) = 7207! = 7(6)(5)(4)(3)(2)(1) = 5040


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Examples3! = 3(2)(1) = 64! = 4(3)(2)(1) = 125! = 5(4)(3)(2)(1) = 1206! = 6(5)(4)(3)(2)(1) = 720

7! = 7(6)(5)(4)(3)(2)(1) = 5040


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Examples3! = 3(2)(1) = 64! = 4(3)(2)(1) = 125! = 5(4)(3)(2)(1) = 1206! = 6(5)(4)(3)(2)(1) = 7207! = 7(6)(5)(4)(3)(2)(1) = 5040


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Definition (Permutation of a Set of Objects)A permutation of a set of distinct objects is an arrangement ofthe objects in a specific order without repetition.


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Theorem (Number of Permutations of n Objects)The number of permutations of n distinct objects withoutrepetition, denoted by Pn,n is

Pn,n = n(n − 1) · · · 2 · 1 = n!


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Example: Now suppose that the director of the art gallerydecides to use only 2 of the 4 available paintings, and they willbe arranged on the wall from left to right. We are now talkingabout a particular arrangement of 2 paintings out of the 4,which is called a permutation of 4 objects taken 2 at a time.

4

×

3

= 12


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4

×

3 = 12


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4×

3 = 12


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4× 3

= 12


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4× 3 = 12


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Definition (Permutation of n Objects Taken r at a Time)A permutation of a set of n distinct objects taken r at a timewithout repetition is an arrangement of r of the n objects in aspecific order.


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Theorem (Number of Permutations of n Objects Taken r at aTime)The number of permutations of n distinct objects taken r at atime without repetition is given by

Pn,r = n(n − 1)(n − 2) · · · (n − r + 1)

orPn,r =

n!(n − r)!

0 ≤ r ≤ n

Note: Pn,n = n!(n−n)! =

n!0! = n! permutations of n objects taken n

at a time.

Note: In place of Pn,r the symbol P(n, r) is often used.


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Example: Find P(5,3)

P(5,3) = (5)(5− 1)(5− 2) = (5)(4)(3) = 60.

This means there are 60 permutations of 5 items taken 3 at atime.Example: Find P(5,5), the number of permutations of 5objects taken 5 at a time.

P(5,5) = 5(4)(3)(2)(1) = 120.


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P(5,3) = (5)(5− 1)(5− 2) = (5)(4)(3) = 60.


P(5,5) = 5(4)(3)(2)(1) = 120.


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P(5,3) = (5)(5− 1)(5− 2) = (5)(4)(3) = 60.

This means there are 60 permutations of 5 items taken 3 at atime.

Example: Find P(5,5), the number of permutations of 5objects taken 5 at a time.

P(5,5) = 5(4)(3)(2)(1) = 120.


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P(5,3) = (5)(5− 1)(5− 2) = (5)(4)(3) = 60.


P(5,5) = 5(4)(3)(2)(1) = 120.


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P(5,3) = (5)(5− 1)(5− 2) = (5)(4)(3) = 60.


P(5,5) = 5(4)(3)(2)(1) = 120.


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Example: A park bench can seat 3 people. How many seatingarrangements are possible if 3 people out of a group of 5 sitdown?

P(5,3) = (5)(4)(3) = 60.


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Example: A park bench can seat 3 people. How many seatingarrangements are possible if 3 people out of a group of 5 sitdown?

P(5,3) = (5)(4)(3) = 60.


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Example: A bookshelf has space for exactly 5 books. Howmany different ways can 5 books be arranged on thisbookshelf?

P(5,5) = 5(4)(3)(2)(1) = 120


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Example: A bookshelf has space for exactly 5 books. Howmany different ways can 5 books be arranged on thisbookshelf?

P(5,5) = 5(4)(3)(2)(1) = 120


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Definition (Combination of n Objects Taken r at a Time)A combination of a set of n distinct objects taken r at a timewithout repetition is an r -element subset of the set of n objects.The arrangement of the elements in the subset does not matter.


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Theorem (Number of Combinations of n Objects Taken r at aTime)The number of combinations of n distinct objects taken r at atime without repetition is given by:

Cn,r =

(nr

)=

Pn,r

r !

=n!

r !(n − r)!0 ≤ r ≤ n

Note: In place of the symbols Cn,r and(

nr

), the symbols

C(n, r) is often used.Jason Aubrey Math 1300 Finite Mathematics

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Examples:

C(8,5) = P(8,5)5! = 8(7)(6)(5)(4)

5(4)(3)(2)(1) = 56

C(8,8) = P(8,8)8! = 8(7)(6)(5)(4)(3)(2)(1)

8(7)(6)(5)(4)(3)(2)(1) = 1


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Examples:

C(8,5) = P(8,5)5! = 8(7)(6)(5)(4)

5(4)(3)(2)(1) = 56

C(8,8) = P(8,8)8! = 8(7)(6)(5)(4)(3)(2)(1)

8(7)(6)(5)(4)(3)(2)(1) = 1


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Example: In how many ways can you choose 5 out of 10friends to invite to a dinner party?

Does the order of selection matter?No! So we use combinations. . .

C(10,5) =P(10,5)

5!=

10(9)(8)(7)(6)5(4)(3)(2)(1)

= 252


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Does the order of selection matter?

No! So we use combinations. . .

C(10,5) =P(10,5)

5!=

10(9)(8)(7)(6)5(4)(3)(2)(1)

= 252


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C(10,5) =P(10,5)

5!=

10(9)(8)(7)(6)5(4)(3)(2)(1)

= 252


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C(10,5) =P(10,5)

5!=

10(9)(8)(7)(6)5(4)(3)(2)(1)

= 252


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Example: How many 5-card poker hands will have 3 aces and2 kings?

The solution involves both the multiplication principle andcombinations.O1: Choose 3 aces out of 4 N1: C4,3O2: Choose 2 kings out of 4 N2: C4,2

Using the multiplication principle, we have:

number of hands = C4,3C4,2

=4!

3!(4− 3)!4!

2!(4− 2)!= 4 · 6 = 24


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The solution involves both the multiplication principle andcombinations.

O1: Choose 3 aces out of 4 N1: C4,3O2: Choose 2 kings out of 4 N2: C4,2



=4!

3!(4− 3)!4!

2!(4− 2)!= 4 · 6 = 24


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=4!

3!(4− 3)!4!

2!(4− 2)!= 4 · 6 = 24


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=4!

3!(4− 3)!4!

2!(4− 2)!= 4 · 6 = 24


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Example: A computer store receives a shipment of 24 laserprinters, including 5 that are defective. Three of these printersare selected to be displayed in the store.

(a) How many selections can be made?

C(24,3) =P(24,3)

3!

=24 · 23 · 22

3 · 2 · 1= 2024


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C(24,3) =P(24,3)

3!

=24 · 23 · 22

3 · 2 · 1= 2024


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C(24,3) =P(24,3)

3!

=24 · 23 · 22

3 · 2 · 1= 2024


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C(24,3) =P(24,3)

3!

=24 · 23 · 22

3 · 2 · 1

= 2024


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C(24,3) =P(24,3)

3!

=24 · 23 · 22

3 · 2 · 1= 2024


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(b) How many of these selections will contain no defectiveprinters?

C(19,3) =P(19,3)

3!

=19 · 18 · 17

3 · 2 · 1= 969


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C(19,3) =P(19,3)

3!

=19 · 18 · 17

3 · 2 · 1= 969


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C(19,3) =P(19,3)

3!

=19 · 18 · 17

3 · 2 · 1

= 969


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C(19,3) =P(19,3)

3!

=19 · 18 · 17

3 · 2 · 1= 969


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Example: Suppose a certain bank has 4 branches inColumbia, 12 branches in St. Louis, and 10 branches inKansas City. The bank must close 6 of these branches.

(a) In how many ways can this be done?

C(26,6) = 230,230

(b) The bank decides to close 2 branches in Columbia, 3branches in St. Louis, and 1 branch in Kansas City. In howmany ways can this be done?O1: pick Columbia branches N1: C(4,2)O2: pick STL branches N2: C(12,3)O3: pick KC branches N3: C(10,1)


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C(26,6) = 230,230



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C(26,6) = 230,230



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C(26,6) = 230,230

(b) The bank decides to close 2 branches in Columbia, 3branches in St. Louis, and 1 branch in Kansas City. In howmany ways can this be done?

O1: pick Columbia branches N1: C(4,2)O2: pick STL branches N2: C(12,3)O3: pick KC branches N3: C(10,1)


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C(26,6) = 230,230



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By the multiplication principle

Number of ways = C(4,2)C(12,3)C(10,1)= 6 · 220 · 10= 13,200


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By the multiplication principle

Number of ways = C(4,2)C(12,3)C(10,1)= 6 · 220 · 10= 13,200


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Example (FS09 Final Exam) During the semester, a professorwrote various problems to possibly include on the final exam ina math course. The course covered 4 chapters of a textbook.She wrote 2 problems from Chapter 1, 3 problems fromChapter 2, 5 problems from Chapter 3, and 4 problems fromChapter 4. She decides to put two questions from each chapteron the final exam.

(a) In how many different ways can she choose two questionsfrom each chapter?


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Example (FS09 Final Exam) During the semester, a professorwrote various problems to possibly include on the final exam ina math course. The course covered 4 chapters of a textbook.She wrote 2 problems from Chapter 1, 3 problems fromChapter 2, 5 problems from Chapter 3, and 4 problems fromChapter 4. She decides to put two questions from each chapteron the final exam.

(a) In how many different ways can she choose two questionsfrom each chapter?


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O1 choose Ch1 questions N1 = C(2,2)O2 choose Ch2 questions N2 = C(3,2)O3 choose Ch3 questions N3 = C(5,2)O4 choose Ch4 questions N4 = C(4,2)

So by the multiplication principle:

C(2,2)C(3,2)C(5,2)C(4,2) = 1(3)(10)(6) = 180


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O1 choose Ch1 questions N1 = C(2,2)O2 choose Ch2 questions N2 = C(3,2)O3 choose Ch3 questions N3 = C(5,2)O4 choose Ch4 questions N4 = C(4,2)

So by the multiplication principle:

C(2,2)C(3,2)C(5,2)C(4,2) = 1(3)(10)(6) = 180


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(b) Suppose she writes the final by (1) choosing a set of 8questions as above, and then (2) randomly ordering those 8questions. Accounting for both operations, how many distinctfinal exams are possible?

O1 choose questions as in (a) N1 = 180O2 put these questions in random order

N2 = P(8,8)

So, by the multiplication principle:

180 · P(8,8) = 180 · 8! = 7,257,600


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O1 choose questions as in (a) N1 = 180O2 put these questions in random order

N2 = P(8,8)


180 · P(8,8) = 180 · 8! = 7,257,600


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O1 choose questions as in (a) N1 = 180O2 put these questions in random order N2 = P(8,8)


180 · P(8,8) = 180 · 8! = 7,257,600


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O1 choose questions as in (a) N1 = 180O2 put these questions in random order N2 = P(8,8)


180 · P(8,8) = 180 · 8! = 7,257,600


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(c) Suppose the questions must be in order by chapter. Forexample, the Chapter 1 questions must come before theChapter 2 questions, the Chapter 2 questions must comebefore the Chapter 3 questions, etc. In this case, how manydifferent final exams are possible?

O1 choose Ch1 questions N1 = P(2,2)O2 choose Ch2 questions N2 = P(3,2)O3 choose Ch3 questions N3 = P(5,2)O4 choose Ch4 questions N4 = P(4,2)


P(2,2)P(3,2)P(5,2)P(4,2) = 2,880


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P(2,2)P(3,2)P(5,2)P(4,2) = 2,880


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P(2,2)P(3,2)P(5,2)P(4,2) = 2,880


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Math 1300: Section 7- 4 Permutations and Combinations