A permutation is an arrangement of a given number of objects, by taking some or all of them at a time. A combination is a selection of a number of objects where the order of the selection is unimportant. Permutations and combinations are defined in terms of the factorial function, which was defined in Chap.4. Recall thatn! =n(n−1)3.2.1.
Principles of Counting
(a) Suppose one operation hasmpossible outcomes and a second operation has n possible outcomes, then the total number of possible outcomes when per- forming thefirst operation followed by the second operation is mn. (Pro- duct Rule).
(b) Suppose one operation hasmpossible outcomes and a second operation has npossible outcomes then the possible outcomes of the first operationorthe second operation is given bym+n. (Sum Rule)
Example (Counting Principle (a)) Suppose a dice is thrown and a coin is then tossed. How many different outcomes are there and what are they?
Solution
There are six possible outcomes from a throw of the dice: 1, 2, 3, 4, 5 or 6, and there are two possible outcomes from the toss of a coin:HorT. Therefore, the total number of outcomes is determined from the product rule as 62 = 12. The outcomes are given by 92 5 Sequences, Series and Permutations and Combinations
1;H
ð Þ;ð2;HÞ;ð3;HÞ;ð4;HÞ;ð5;HÞ;ð6;HÞ;ð1;TÞ;ð2;TÞ;ð3;TÞ;ð4;TÞ;ð5;TÞ;ð6;TÞ Example (Counting Principle (b)) Suppose a dice is thrown and if the number is even a coin is tossed and if it is odd then there is a second throw of the dice. How many different outcomes are there?
Solution
There are two experiments involved with the first experiment involving an even number and a toss of a coin. There are three possible outcomes that result in an even number and two outcomes from the toss of a coin. Therefore, there are 32 = 6 outcomes from thefirst experiment.
The second experiment involves an odd number from the throw of a dice and the further throw of the dice. There are three possible outcomes that result in an odd number and six outcomes from the throw of a dice. Therefore, there are 36 = 18 outcomes from the second experiment.
Finally, there are six outcomes from thefirst experiment and 18 outcomes from the second experiment, and so from the sum rule there are a total of 6 + 18 = 24 outcomes.
Pigeonhole Principle
The pigeonhole principle states that if nitems are placed into mcontainers (with n> m) then at least one container must contain more than one item.
Examples (Pigeonhole Principle)
(a) Suppose there is a group of 367 people then there must be at least two people with the same birthday.
This is clear as there are 365 days in a year (with 366 days in a leap year), and so as there are at most 366 possible birthdays in a year. The group size is 367 people, and so there must be at least two people with the same birthday.
(b) Suppose that a class of 102 students are assessed in an examination (the outcome from the exam is a mark between 0 and 100). Then, there are at least two students who receive the same mark.
This is clear as there are 101 possible outcomes from the test (as the mark that a student may achieve is between is between 0 and 100), and as there are 102 students in the class and 101 possible outcomes from the test, then there must be at least two students who receive the same mark.
Permutations
A permutation is an arrangement of a number of objects in a definite order.
Consider the three letters A, B and C. If these letters are written in a row then there are six possible arrangements:
ABC or ACB or BAC or BCA or CAB or CBA
There is a choice of three letters for thefirst place, then there is a choice of two letters for the second place, and there is only one choice for the third place.
Therefore, there are 32 1 = 6 arrangements.
If there arendifferent objects to arrange then the total number of arrangements (permutations) ofn objects is given byn! =n(n −1)(n − 2)…3.2.1.
Consider the four letters A, B, C and D. How many arrangements (taking two letters at a time with no repetition) of these letters can be made?
There are four choices for thefirst letter and three choices for the second letter, and so there are 12 possible arrangements. These are given by
AB or AC or AD or BA or BC or BD or CA or CB or CD or DA or DB or DC The total number of arrangements ofndifferent objects takingrat a time (rn) is given bynPr= n(n− 1)(n −2)…(n− r+ 1). It may also be written as
nPr¼ n!
ðnrÞ!
Example (Permutations) Suppose A, B, C, D, E and F are six students. How many ways can they be seated in a row if
(i) There is no restriction on the seating.
(ii) A and B must sit next to one another (iii) A and B must not sit next to one another Solution
For unrestricted seating the number of arrangements is given by 6.5.4.3.2.1 = 6! = 720.
For the case where A and B must be seated next to one another, then consider A and B as one person, and then thefive people may be arranged in 5! = 120 ways.
There are 2! = 2 ways in which AB may be arranged, and so there are 2!
5! = 240 arrangements.
AB C D E F
For the case where A and B must not be seated next to one another, then this is given by the difference between the total number of arrangements and the number of arrangements with A and B together: i.e. 720– 240 = 480.
94 5 Sequences, Series and Permutations and Combinations
Combinations
A combination is a selection of a number of objects in any order, and the order of the selection is unimportant, in that both AB and BA represent the same selection.
The total number of arrangements ofndifferent objects takingrat a time is given bynPr, and we can determine that the number of ways thatrobjects can be selected fromndifferent objects from this, as each selection may be permutedr! times, and so the total number of selections is r! total number of combinations. That is,
nPr =r!nCr, and we may also write this as
ðnrÞ ¼ n!
r!ðnrÞ!¼nðn1Þ. . .ðnrþ1Þ r!
It is clear from the definition that
ðnrÞ ¼ ðnnrÞ
Example 1 (Combinations) How many ways are there to choose a team of 11 players from a panel of 15 players?
Solution
Clearly, the number of ways is given byð1511Þ ¼ ð154 Þ That is, 15.14.13.12/4.3.2.1 = 1365.
Example 2 (Combinations) How many ways can a committee of four people be chosen from a panel of 10 people where
(i) There is no restriction on membership of the panel.
(ii) A certain person must be a member.
(iii) A certain person must not be a member.
Solution
For (i) with no restrictions on membership the number of selections of a committee of four people from a panel of 10 people is given byð104 Þ ¼210
For (ii) where one person must be a member of the committee then this involves choosing three people from a panel of nine people and is given byð93Þ ¼84
For (iii) where one person must not be a member of the committee then this involves choosing four people from a panel of nine people, and is given by ð94Þ ¼126