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Unit 7 Permutation and Combination IT Disicipline ITD1111 Discrete Mathematics & Statistics STDTLP 1 Unit 7 Permutation and Combination
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IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP2 In this section, techniques will be introduced for counting the unordered selections of distinct objects and the ordered arrangements of objects of a finite set.
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP3 7.1 Arrangements The number of ways of arranging n unlike objects in a line is n !. Note: n ! = n (n-1) (n-2) ···3 x 2 x 1
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP4 Example 7.1-1 It is known that the password on a computer system contain the three letters A, B and C followed by the six digits 1, 2, 3, 4, 5, 6. Find the number of possible passwords.
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP5 Solution 7.1-1 There are 3! ways of arranging the letters A, B and C, and 6! ways of arranging the digits 1, 2, 3, 4, 5, 6. Therefore the total number of possible passwords is 3! x 6! = 4320. i.e. 4320 different passwords can be formed.
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP6 Like Objects The number of ways of arranging in a line n objects, of which p are alike, is
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP7 The result can be extended as follows: The number of ways of arranging in a line n objects of which p of one type are alike, q of a second type are alike, r of a third type are alike, and so on, is
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP8 Example 7.1-2 Find the number of ways that the letters of the word STATISTICS can be arranged.
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP9 Solution 7.1-2 The word STATISTICS contains 10 letters, in which S occurs 3 times, T occurs 3 times and I occurs twice.
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP10 Therefore the number of ways is That is, there are 50400 ways of arranging the letter in the word STATISTICS.
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP11 Example 7.1-3 A six-digit number is formed from the digits 1, 1, 2, 2, 2, 5 and repetitions are not allowed. How many these six-digit numbers are divisible by 5?
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP12 Solution 7.1-3 If the number is divisible by 5 then it must end with the digit 5. Therefore the number of these six-digit numbers which are divisible by 5 is equal to the number of ways of arranging the digits 1, 1, 2, 2, 2.
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP13 Then, the required number is That is, there are 10 of these six-digit numbers are divisible by 5.
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP14 7.2 Permutations A permutation of a set of distinct objects is an ordered arrangement of these objects. An ordered arrangement of r elements of a set is called an r-permutation. The number of r-permutations of a set with n distinct elements,
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP15 Note: 0! is defined to 1, so i.e. the number of permutations of r objects taken from n unlike objects is:
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP16 Example 7.2-1 Find the number of ways of placing 3 of the letters A, B, C, D, E in 3 empty spaces.
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP17 Solution 7.2-1 The first space can be filled in 5 ways. The second space can be filled in 4 ways. The third space can be filled in 3 ways.
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP18 Therefore there are 5 x 4 x 3 ways of arranging 3 letters taken from 5 letters. This is the number of permutations of 3 objects taken from 5 and it is written as P(5, 3), so P(5, 3) = 5 x 4 x 3 = 60.
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP19 On the other hand, 5 x 4 x 3 could be written as Notice that the order in which the letters are arranged is important --- ABC is a different permutation from ACB.
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP20 Example 7.2-2 How many different ways are there to select one chairman and one vice chairman from a class of 20 students.
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP21 Solution 7.2-2 The answer is given by the number of 2-permutations of a set with 20 elements. This is P(20, 2) = 20 x 19 = 380
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP22 7.3 Combinations An r-combination of elements of a set is an unordered selection of r elements from the set. Thus, an r-combination is simply a subset of the set with r elements.
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP23 The number of r-combinations of a set with n elements, where n is a positive integer and r is an integer with 0 <= r <= n, i.e. the number of combinations of r objects from n unlike objects is
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP24 Example 7.3-1 How many different ways are there to select two class representatives from a class of 20 students?
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP25 Solution 7.3-1 The answer is given by the number of 2- combinations of a set with 20 elements. The number of such combinations is
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP26 Example 7.3-2 A committee of 5 members is chosen at random from 6 faculty members of the mathematics department and 8 faculty members of the computer science department.
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP27 In how many ways can the committee be chosen if (a)there are no restrictions; (b)there must be more faculty members of the computer science department than the faculty members of the mathematics department.
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP28 Solution 7.3-2 (a)There are 14 members, from whom 5 are chosen. The order in which they are chosen is not important. So the number of ways of choosing the committee is C(14, 5) = 2002.
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP29 (b)If there are to be more faculty members of the computer science department than the faculty members of the mathematics department, then the following conditions must be fulfilled.
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP30 (i)5 faculty members of the computerscience department. The number of ways of choosing is C(8, 5) = 56. (ii)4 faculty members of the computer science department and 1 faculty member of the mathematics department
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP31 The number of ways of choosing is C(8, 4) x C(6, 1) = 70 x 6 = 420. (iii) 3 faculty members of the computer science department and 2 faculty members of the mathematics department The number of ways of choosing is C(8, 3) x C(6, 2) = 56 x 15 = 840
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Unit 7 Permutation and Combination IT DisiciplineITD1111 Discrete Mathematics & Statistics STDTLP32 Therefore the total number of ways of choosing the committee is 56 + 420 + 840 = 1316.
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