Discrete Mathematics
WEN-CHING LIEN Department of Mathematics National Cheng Kung University
2008
1.2: Permutation
Definition (1.1)
For an integer n≥0, n factorial(denoted n!) is defined by 0! =1
n! = (n)(n−1)(n−2) · · · (3)(2)(1), for n ≥1.
Definition (1.2)
Given a collection of n distinct objects, any (linear) arrangement
1.2: Permutation
Definition (1.1)
For an integer n≥0, n factorial(denoted n!) is defined by 0! =1
n! = (n)(n−1)(n−2) · · · (3)(2)(1), for n ≥1.
Definition (1.2)
Given a collection of n distinct objects, any (linear) arrangement of these objects is called a permutation of the collection.
If there are n distinct objects and r is an integer, with 1≤r ≤n, then by the rule of product, the number of permutations of size r for the n objects is
P(n,r) = n×(n−1) × (n−2) × · · · × (n−r +1)
= n(n−1)(n−2) · · · (n−r +1) × (n−r)(n−r −1) · · · (3)(2)(1) (n−r)(n−r −1) · · · (3)(2)(1)
= n!
(n−r)!
Example (1.10)
The number of permutations of the letters in the word
COMPUTER is8!.If only five of the letters are used, the number of permutations (of size 5) is
P(8,5) =8!/(8−5)! =8!/3! =6720.If repetitions of letters are allowed,the number of possible 12-letter sequences is812 .
= 6.872×1010
Example (1.10)
The number of permutations of the letters in the word
COMPUTER is 8!.If only five of the letters are used, the number of permutations (of size 5) is
P(8,5) =8!/(8−5)! =8!/3! =6720.If repetitions of letters are allowed,the number of possible 12-letter sequences is 812 .
= 6.872×1010
Example (1.10)
The number of permutations of the letters in the word
COMPUTER is 8!.If only five of the letters are used, the number of permutations (of size 5) is
P(8,5) =8!/(8−5)! =8!/3! =6720.If repetitions of letters are allowed,the number of possible 12-letter sequences is812 .
= 6.872×1010
Example (1.10)
The number of permutations of the letters in the word
COMPUTER is 8!.If only five of the letters are used, the number of permutations (of size 5) is
P(8,5) =8!/(8−5)! =8!/3! =6720.If repetitions of letters are allowed,the number of possible 12-letter sequences is 812 .
= 6.872×1010
Rule:
If there are n objects with n1indistinguishable objects of a first type, n2indistinguishable objects of a second type,· · · ,and nr indistinguishable objects of an r th type,
wheren1+n2+ · · · +nr =n, then there are n n!
1!n2!···nr! (linear) arrangements of the given n objects.
Example:[1.13]
The MASSASAUGA is a brown and white venomous snake indigenous to North America. Arranging all of the letters in MASSASAUGA,we find that there are
10!
4!3!1!1!1! =25200 possible arrangements. Among these are
7!
3!1!1!1!1! =840
in which all four A’s are together. To get this last result, we considered all arrangements of the seven symbols AAAA(one
Example:[1.13]
The MASSASAUGA is a brown and white venomous snake indigenous to North America. Arranging all of the letters in MASSASAUGA,we find that there are
10!
4!3!1!1!1! =25200 possible arrangements. Among these are
7!
3!1!1!1!1! =840
in which all four A’s are together. To get this last result, we considered all arrangements of the seven symbols AAAA(one symbol),S, S, S, M, U, G.
Example:[1.13]
The MASSASAUGA is a brown and white venomous snake indigenous to North America. Arranging all of the letters in MASSASAUGA,we find that there are
10!
4!3!1!1!1! =25200 possible arrangements. Among these are
7!
3!1!1!1!1! =840
in which all four A’s are together. To get this last result, we considered all arrangements of the seven symbols AAAA(one
