4 3 permutations
TRANSCRIPT
PERMUTATIONS and COMBINATIONSCOUNTING
Dr. Felix P. Muga IIMathematics Department
School of Science and EngineeringAteneo de Manila University
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 1 / 21
Introduction
A permutation of a set of distinct objects is an ordered arrangement ofthese objects.An ordered arrangement of r elements of a set is called anr -permutation.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 2 / 21
Introduction
A permutation of a set of distinct objects is an ordered arrangement ofthese objects.An ordered arrangement of r elements of a set is called anr -permutation.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 2 / 21
ExampleLet S = {1, 2, 3}. The arrangement of 3, 1, 2 is a permutation of S.
The arrangement 3, 1 is a 2-permutation of S.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 3 / 21
ExampleLet S = {1, 2, 3}. The arrangement of 3, 1, 2 is a permutation of S.
The arrangement 3, 1 is a 2-permutation of S.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 3 / 21
TheoremThe number of r -permutations of a set with n distinct elements is
P(n, r) = n(n − 1)(n − 2) · · · (n − r + 1)
P(n, r) =n!
(n − r)!
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 4 / 21
TheoremThe number of r -permutations of a set with n distinct elements is
P(n, r) = n(n − 1)(n − 2) · · · (n − r + 1)
P(n, r) =n!
(n − r)!
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 4 / 21
ExampleHow many ways are there to select a first-prize winner, a second-prizewinner, and a third-prize winner from 100 different people who haveentered a contest?
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 5 / 21
ExampleSuppose that there are 8 runners in a race. The winner receives a goldmedal, the second-place finisher a silver medal, and the third-placefinisher receives a bronze medal. How many different ways are thereto award these medals, if all possible outcomes of the race can occurand there are no ties?
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 6 / 21
ExampleSuppose that a saleswoman has to visit eight different cities. She mustbegin her trip in a specified city, but she can visit the other seven citiesin any order she wishes. How many possible orders can thesaleswoman use when visiting these cities?
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 7 / 21
ExampleHow many permutations of the letters ABCDEFGH contain the stringsABC?
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 8 / 21
Combinations
An r -combination of elements of a set is an unordered selection of relements from the set.
An r -combination is simply a subset of the set with r elements.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 9 / 21
Combinations
An r -combination of elements of a set is an unordered selection of relements from the set.
An r -combination is simply a subset of the set with r elements.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 9 / 21
ExampleLet S be the set {1, 2, 3, 4}. Then {1, 3, 4} is a 3-combination of S.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 10 / 21
Binomial Coefficient
C(n, r) denotes the number of r -combinations of a set with n distinctelements.
C(n, r) =(n
r
)and is called a binomial coefficient.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 11 / 21
ExampleC(4, 2) = 6, since the 2-combinations of {a, b, c, d} are the six subsets
{a, b}, {a, c}, {a, d},{b, c}, {b, d}, {c, d}.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 12 / 21
TheoremThe number of r -combinations of a set with n elements, where n is anonnegative integer and r is an integer with 0 ≤ r ≤ n, equals
C(n, r) =n!
r !(n − r)!.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 13 / 21
CorollaryLet n and r be nonnegative integers with r ≤ n. Then
C(n, r) = C(n, n − r).
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 14 / 21
ExampleHow many ways are there to select three contestants from a10-member programming team?
Solution.C(10, 3) =
10!
3!7!= 120.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 15 / 21
ExampleHow many ways are there to select three contestants from a10-member programming team?
Solution.C(10, 3) =
10!
3!7!= 120.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 15 / 21
ExampleHow many bit strings of length 20 contains exactly twelve 1s?
Solution.C(20, 12) =
20!
12!7!= 125, 970.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 16 / 21
ExampleHow many bit strings of length 20 contains exactly twelve 1s?
Solution.C(20, 12) =
20!
12!7!= 125, 970.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 16 / 21
ExampleA club has 25 members.
1 How many ways are there to choose four members of the club toserve on an executive committee?Solution.C(25, 4) = 12, 650.
2 How many ways are there to choose a president, vice-president,secretary and treasurer of the club?Solution.P(25, 4) = 303, 600.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 17 / 21
ExampleA club has 25 members.
1 How many ways are there to choose four members of the club toserve on an executive committee?
Solution.C(25, 4) = 12, 650.
2 How many ways are there to choose a president, vice-president,secretary and treasurer of the club?Solution.P(25, 4) = 303, 600.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 17 / 21
ExampleA club has 25 members.
1 How many ways are there to choose four members of the club toserve on an executive committee?Solution.C(25, 4) = 12, 650.
2 How many ways are there to choose a president, vice-president,secretary and treasurer of the club?Solution.P(25, 4) = 303, 600.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 17 / 21
ExampleA club has 25 members.
1 How many ways are there to choose four members of the club toserve on an executive committee?Solution.C(25, 4) = 12, 650.
2 How many ways are there to choose a president, vice-president,secretary and treasurer of the club?
Solution.P(25, 4) = 303, 600.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 17 / 21
ExampleA club has 25 members.
1 How many ways are there to choose four members of the club toserve on an executive committee?Solution.C(25, 4) = 12, 650.
2 How many ways are there to choose a president, vice-president,secretary and treasurer of the club?Solution.P(25, 4) = 303, 600.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 17 / 21
ExampleHow many bit strings contain exactly eight 0s and ten 1s if every 0must be immediately followed by a 1?
Solution.We have a set of 10 elements where 8 elements of the same kindwhich is a 01 and 2 elements of the same kind which is a 1.Hence, there are C(10, 8) = C(10, 2) = 45 strings satisfying the givencondition.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 18 / 21
ExampleHow many bit strings contain exactly eight 0s and ten 1s if every 0must be immediately followed by a 1?
Solution.We have a set of 10 elements where 8 elements of the same kindwhich is a 01 and 2 elements of the same kind which is a 1.Hence, there are C(10, 8) = C(10, 2) = 45 strings satisfying the givencondition.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 18 / 21
Example
How many bit strings of length 10 contain at least three 1s and at least three 0s?
Solution.
Case 1. There are exactly three 1s and exactly seven 0s.
C(10, 3) = C(10, 7) = 120.
Case 2. There are exactly four 1s and exactly six 0s.
C(10, 4) = C(10, 6) = 210.
Case 3. There are exactly five 1s and exactly five 0s.
C(10, 5) = 252.
Case 4. There are exactly six 1s and exactly four 0s.
C(10, 6) = C(10, 4) = 210.
Case 5. There are exactly seven 1s and exactly three 0s.
C(10, 7) = C(10, 3) = 120.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 19 / 21
Example
How many bit strings of length 10 contain at least three 1s and at least three 0s?
Solution.
Case 1. There are exactly three 1s and exactly seven 0s.
C(10, 3) = C(10, 7) = 120.
Case 2. There are exactly four 1s and exactly six 0s.
C(10, 4) = C(10, 6) = 210.
Case 3. There are exactly five 1s and exactly five 0s.
C(10, 5) = 252.
Case 4. There are exactly six 1s and exactly four 0s.
C(10, 6) = C(10, 4) = 210.
Case 5. There are exactly seven 1s and exactly three 0s.
C(10, 7) = C(10, 3) = 120.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 19 / 21
Example
How many bit strings of length 10 contain at least three 1s and at least three 0s?
Solution.
Case 1. There are exactly three 1s and exactly seven 0s.
C(10, 3) = C(10, 7) = 120.
Case 2. There are exactly four 1s and exactly six 0s.
C(10, 4) = C(10, 6) = 210.
Case 3. There are exactly five 1s and exactly five 0s.
C(10, 5) = 252.
Case 4. There are exactly six 1s and exactly four 0s.
C(10, 6) = C(10, 4) = 210.
Case 5. There are exactly seven 1s and exactly three 0s.
C(10, 7) = C(10, 3) = 120.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 19 / 21
Example
How many bit strings of length 10 contain at least three 1s and at least three 0s?
Solution.
Case 1. There are exactly three 1s and exactly seven 0s.
C(10, 3) = C(10, 7) = 120.
Case 2. There are exactly four 1s and exactly six 0s.
C(10, 4) = C(10, 6) = 210.
Case 3. There are exactly five 1s and exactly five 0s.
C(10, 5) = 252.
Case 4. There are exactly six 1s and exactly four 0s.
C(10, 6) = C(10, 4) = 210.
Case 5. There are exactly seven 1s and exactly three 0s.
C(10, 7) = C(10, 3) = 120.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 19 / 21
Example
How many bit strings of length 10 contain at least three 1s and at least three 0s?
Solution.
Case 1. There are exactly three 1s and exactly seven 0s.
C(10, 3) = C(10, 7) = 120.
Case 2. There are exactly four 1s and exactly six 0s.
C(10, 4) = C(10, 6) = 210.
Case 3. There are exactly five 1s and exactly five 0s.
C(10, 5) = 252.
Case 4. There are exactly six 1s and exactly four 0s.
C(10, 6) = C(10, 4) = 210.
Case 5. There are exactly seven 1s and exactly three 0s.
C(10, 7) = C(10, 3) = 120.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 19 / 21
Example
How many bit strings of length 10 contain at least three 1s and at least three 0s?
Solution.
Case 1. There are exactly three 1s and exactly seven 0s.
C(10, 3) = C(10, 7) = 120.
Case 2. There are exactly four 1s and exactly six 0s.
C(10, 4) = C(10, 6) = 210.
Case 3. There are exactly five 1s and exactly five 0s.
C(10, 5) = 252.
Case 4. There are exactly six 1s and exactly four 0s.
C(10, 6) = C(10, 4) = 210.
Case 5. There are exactly seven 1s and exactly three 0s.
C(10, 7) = C(10, 3) = 120.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 19 / 21
Example
How many bit strings of length 10 contain at least three 1s and at least three 0s?
Solution.
Case 1. There are exactly three 1s and exactly seven 0s.
C(10, 3) = C(10, 7) = 120.
Case 2. There are exactly four 1s and exactly six 0s.
C(10, 4) = C(10, 6) = 210.
Case 3. There are exactly five 1s and exactly five 0s.
C(10, 5) = 252.
Case 4. There are exactly six 1s and exactly four 0s.
C(10, 6) = C(10, 4) = 210.
Case 5. There are exactly seven 1s and exactly three 0s.
C(10, 7) = C(10, 3) = 120.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 19 / 21
Example
How many bit strings of length 10 contain at least three 1s and at least three 0s?
Solution.
Case 1. There are exactly three 1s and exactly seven 0s.
C(10, 3) = C(10, 7) = 120.
Case 2. There are exactly four 1s and exactly six 0s.
C(10, 4) = C(10, 6) = 210.
Case 3. There are exactly five 1s and exactly five 0s.
C(10, 5) = 252.
Case 4. There are exactly six 1s and exactly four 0s.
C(10, 6) = C(10, 4) = 210.
Case 5. There are exactly seven 1s and exactly three 0s.
C(10, 7) = C(10, 3) = 120.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 19 / 21
Example
How many bit strings of length 10 contain at least three 1s and at least three 0s?
Solution.
Case 1. There are exactly three 1s and exactly seven 0s.
C(10, 3) = C(10, 7) = 120.
Case 2. There are exactly four 1s and exactly six 0s.
C(10, 4) = C(10, 6) = 210.
Case 3. There are exactly five 1s and exactly five 0s.
C(10, 5) = 252.
Case 4. There are exactly six 1s and exactly four 0s.
C(10, 6) = C(10, 4) = 210.
Case 5. There are exactly seven 1s and exactly three 0s.
C(10, 7) = C(10, 3) = 120.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 19 / 21
Example
How many bit strings of length 10 contain at least three 1s and at least three 0s?
Solution.
Case 1. There are exactly three 1s and exactly seven 0s.
C(10, 3) = C(10, 7) = 120.
Case 2. There are exactly four 1s and exactly six 0s.
C(10, 4) = C(10, 6) = 210.
Case 3. There are exactly five 1s and exactly five 0s.
C(10, 5) = 252.
Case 4. There are exactly six 1s and exactly four 0s.
C(10, 6) = C(10, 4) = 210.
Case 5. There are exactly seven 1s and exactly three 0s.
C(10, 7) = C(10, 3) = 120.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 19 / 21
Example
How many bit strings of length 10 contain at least three 1s and at least three 0s?
Solution.
Case 1. There are exactly three 1s and exactly seven 0s.
C(10, 3) = C(10, 7) = 120.
Case 2. There are exactly four 1s and exactly six 0s.
C(10, 4) = C(10, 6) = 210.
Case 3. There are exactly five 1s and exactly five 0s.
C(10, 5) = 252.
Case 4. There are exactly six 1s and exactly four 0s.
C(10, 6) = C(10, 4) = 210.
Case 5. There are exactly seven 1s and exactly three 0s.
C(10, 7) = C(10, 3) = 120.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 19 / 21
Example
How many bit strings of length 10 contain at least three 1s and at least three 0s?
Solution.
Case 1. There are exactly three 1s and exactly seven 0s.
C(10, 3) = C(10, 7) = 120.
Case 2. There are exactly four 1s and exactly six 0s.
C(10, 4) = C(10, 6) = 210.
Case 3. There are exactly five 1s and exactly five 0s.
C(10, 5) = 252.
Case 4. There are exactly six 1s and exactly four 0s.
C(10, 6) = C(10, 4) = 210.
Case 5. There are exactly seven 1s and exactly three 0s.
C(10, 7) = C(10, 3) = 120.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 19 / 21
Example
How many bit strings of length 10 contain at least three 1s and at least three 0s?
Continuation.There are 120 + 210 + 252 + 210 + 120 = 912 strings containing at least three 1s andat least three 0s.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 20 / 21
Example
How many bit strings of length 10 contain at least three 1s and at least three 0s?
Continuation.There are 120 + 210 + 252 + 210 + 120 = 912 strings containing at least three 1s andat least three 0s.
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 20 / 21
ExampleLet S be a set with n distinct elements. How many subsets does Shave including the empty set and itself?
Solution.
|P(S)| = 2n.
However, |P(S)| = C(n, 0) + C(n, 1) + · · ·+ C(n, n − 1) + C(n, n).
Thus, 2n = C(n, 0) + C(n, 1) + · · ·+ C(n, n − 1) + C(n, n).
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 21 / 21
ExampleLet S be a set with n distinct elements. How many subsets does Shave including the empty set and itself?
Solution.
|P(S)| = 2n.
However, |P(S)| = C(n, 0) + C(n, 1) + · · ·+ C(n, n − 1) + C(n, n).
Thus, 2n = C(n, 0) + C(n, 1) + · · ·+ C(n, n − 1) + C(n, n).
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 21 / 21
ExampleLet S be a set with n distinct elements. How many subsets does Shave including the empty set and itself?
Solution.
|P(S)| = 2n.
However, |P(S)| = C(n, 0) + C(n, 1) + · · ·+ C(n, n − 1) + C(n, n).
Thus, 2n = C(n, 0) + C(n, 1) + · · ·+ C(n, n − 1) + C(n, n).
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 21 / 21
ExampleLet S be a set with n distinct elements. How many subsets does Shave including the empty set and itself?
Solution.
|P(S)| = 2n.
However, |P(S)| = C(n, 0) + C(n, 1) + · · ·+ C(n, n − 1) + C(n, n).
Thus, 2n = C(n, 0) + C(n, 1) + · · ·+ C(n, n − 1) + C(n, n).
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 21 / 21
ExampleLet S be a set with n distinct elements. How many subsets does Shave including the empty set and itself?
Solution.
|P(S)| = 2n.
However, |P(S)| = C(n, 0) + C(n, 1) + · · ·+ C(n, n − 1) + C(n, n).
Thus, 2n = C(n, 0) + C(n, 1) + · · ·+ C(n, n − 1) + C(n, n).
Felix P. Muga II () PERMUTATIONS and COMBINATIONS AMC 124 21 / 21