     # Permutation and Combinations (1)

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• Working with Sets There are two ways to solve problems involving overlapping sets

1. Tabulation 2. Venn Diagrams

Ill go through both quickly here with some examples. Tabulation Problems that involve 2 or more given sets of data that partially intersect each other are termed overlapping sets. Problems that involve 2 sets can be solved using a table as shown below:

B !B Total A !A

Total The example below shows how you can utilize this method. Example 1: (Question taken off GMATClub Math Forum) Of the students at a certain school, 50% of the students are in class A, 30% of the students are in class B, and 20% of the students are in class C. Of the students in class A, 20% are wearing eyeglasses. Of the students in class B, 10% are wearing eyeglasses. Of the students in class C, 10% are wearing eyeglasses. If a student is selected at random, what is the probability that the student is in class B not wearing eyeglasses?

Eye Glass No Eyeglass Total A 10 40 50 B 3 27 30 C 2 18 20

Total 15 85 100 Since were working with percentages, one good method is to work with a total of 100 students. 50% are in class A, so 50 students are in class A 30% are in class B, so 30 students are in class B 20% are in class C, so 20 students are in class C We fill in the values under the total column, since these numbers represent the total number of students for each individual class.

• Now, were also told in class A, 20% wear eyeglasses. Therefore, 10 students in class A wear eyeglasses. Also, were told 10% of the students in class B wear eyeglasses. Therefore, 3 students in class B wear eyeglasses. Finally, were told 10% of the students in class C wear eyeglasses. Therefore, 2 students in class C wear eyeglasses. We can now fill in these values under the column eye glasses for each individual class. We can also sum up the total number of people who wear eyeglasses (10+3+2 = 15) Now all we need to do is to fill up the remaining spaces in the table. Number of students in Class A who do not wear eye glasses = 50-10 = 40 Number of students in Class B who do not wear eye glasses = 30-3 = 27 Number of students in Class C who do not wear eye glasses = 20-2 = 18 Therefore total number of students who do not wear eyeglasses = 40+27+18 = 85 (Note: You can also calculate the number of students who do not wear eyeglasses by subtracting 15 from 100) Since were asked for the probability of a student from class B not wearing eyeglasses, we will need the total number of students who do not wear eyeglasses (85) and the number of students in class B who do not wear eyeglasses (27) The probability is therefore 27/85 Example 2 30 people are in a room. 20 of them play golf. 15 of them play golf and tennis. If everyone plays at least one sport, how many of the people play tennis only? Again, we can tabulate the values in order to solve the problem.

Plays Tennis Do Not Play Tennis Total Plays Golf 15 5 20

Do Not Play Golf 10 0 10 Total 25 5 30

Set the value of total to 30 (since 30 people are in the room) Set the value of plays golf to 20 (since 20 of them plays golf) Set the value of 15 to the intersection of plays golf and plays tennis (since 15 of them play both)

• We also know that everyone plays at least one sport. So the intersection of do not play golf and do not play tennis should be 0. We can now proceed to fill in the rest: Number of people who do not plays golf = 30-20 = 10 Number of people who plays golf but do not plays tennis = 20-15 = 5 Total number of people who do not play tennis = 5+0=5 Total number of people who plays tennis = 30-5 = 25 Number of people who do not play golf but plays tennis = 25-15=10 So now we know how many people play tennis only: = 10 Venn Diagrams Problems involving 3 overlapping sets can be solved using a Venn diagram. Venn diagrams should be worked from inside out.

A

B

C D

E

F G

Venn diagrams should be worked from inside out. So A should be filled in first, followed by B, C and D, and finally E, F and G. The example below will illustrate how to solve a problem involving 3 overlapping sets using a Venn diagram.

• Example 1 (Question taken off GMATClub Math Forum) Of the 60 students of the IT class of 2003 at UC Berkeley, 30 use a Mac, 20 use a SPARC workstation, and 20 use a SGI. 13 use both Mac and SPARC, 5 uses both MAC and SGI, and 8 uses both SGI and SPARC. If 5 of the students use all three, how many don't use any of the three (not a Mac, not a SPARC, not a SGI, but maybe a crappy Intel machine)? 0 5 10 11 12 Using the Venn diagram, we can now proceed to fill in the values. 5 students use all three machines, so A = 5 13 students use both Mac and SPARC, so C = 13-5 = 8 5 students use both Mac and SGI, so B = 5-5=0 8 students use both SGI and SPARC, so D=8-5 =3 30 students use a MAC, so F = 30-8-5=17 20 students use a SPARC, so E = 20-8-5-3=4 20 students use a SGI, so G = 20-5-3=12 So now your Venn diagram should look this way:

• 5

0

8 3

4

17 12Mac

SPARC

SGI

Use none of 3

To solve for the number of students who dont use either of these machines, take 60 and subtract off from this number the sum of all the numbers in the Venn diagram. So number of students who use none of the 3 computers is: 60-(17+8+5+4+3+12) = 11 So Answer = D

• 138 MATHEMATICS

The number of ways is 5 4 3 = 60.Continuing the same way, we find that

The number of 4 flag signals = 5 4 3 2 = 120and the number of 5 flag signals = 5 4 3 2 1 = 120Therefore, the required no of signals = 20 + 60 + 120 + 120 = 320.

EXERCISE 7.1

1. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5assuming that(i) repetition of the digits is allowed?(ii) repetition of the digits is not allowed?

2. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if thedigits can be repeated?

3. How many 4-letter code can be formed using the first 10 letters of the Englishalphabet, if no letter can be repeated?

4. How many 5-digit telephone numbers can be constructed using the digits 0 to 9 ifeach number starts with 67 and no digit appears more than once?

5. A coin is tossed 3 times and the outcomes are recorded. How many possibleoutcomes are there?

6. Given 5 flags of different colours, how many different signals can be generated ifeach signal requires the use of 2 flags, one below the other?

7.3 PermutationsIn Example 1 of the previous Section, we are actually counting the different possiblearrangements of the letters such as ROSE, REOS, ..., etc. Here, in this list, eacharrangement is different from other. In other words, the order of writing the letters isimportant. Each arrangement is called a permutation of 4 different letters taken allat a time. Now, if we have to determine the number of 3-letter words, with or withoutmeaning, which can be formed out of the letters of the word NUMBER, where therepetition of the letters is not allowed, we need to count the arrangements NUM,NMU, MUN, NUB, ..., etc. Here, we are counting the permutations of 6 differentletters taken 3 at a time. The required number of words = 6 5 4 = 120 (by usingmultiplication principle).

If the repetition of the letters was allowed, the required number of words wouldbe 6 6 6 = 216.

• PERMUTATIONS AND COMBINATIONS 139

Definition 1 A permutation is an arrangement in a definite order of a number ofobjects taken some or all at a time.

In the following sub Section, we shall obtain the formula needed to answer thesequestions immediately.

7.3.1 Permutations when all the objects are distinct

Theorem 1 The number of permutations of n different objects taken r at a time,where 0 < r n and the objects do not repeat is n ( n 1) ( n 2). . .( n r + 1),which is denoted by nPr.

Proof There will be as many permutations as there are ways of filling in r vacant

places . . . by

r vacant places the n objects. The first place can be filled in n ways; following which, the second placecan be filled in (n 1) ways, following which the third place can be filled in (n 2)ways,..., the rth place can be filled in (n (r 1)) ways. Therefore, the number ofways of filling in r vacant places in succession is n(n 1) (n 2) . . . (n (r 1)) orn ( n 1) (n 2) ... (n r + 1)

This expression for nPr is cumbersome and we need a notation which will help toreduce the size of this expression. The symbol n! (read as factorial n or n factorial )comes to our rescue. In the following text we will learn what actually n! means.

7.3.2 Factorial notation The notation n! represents the product of first n naturalnumbers, i.e., the product 1 2 3 . . . (n 1) n is denoted as n!. We read thissymbol as n factorial. Thus, 1 2 3 4 . . . (n 1) n = n !

1 = 1 !1 2 = 2 !1 2 3 = 3 !1 2 3 4 = 4 ! and so on.

We define 0 ! = 1We can write 5 ! = 5 4 ! = 5 4 3 ! = 5 4 3 2 !

= 5 4 3 2 1!Clearly, for a natural number n

n ! = n (n 1) != n (n 1) (n 2) ! [provided (n 2)]= n (n 1) (n 2) (n 3) ! [provided (n 3)]

and so on.

• 140 MATHEMATICS

Example 5 Evaluate (i) 5 ! (ii) 7 ! (iii) 7 ! 5!

Solution (i) 5 ! = 1 2 3 4 5 = 120(ii) 7 ! = 1 2 3 4 5 6 7 = 5040

and (iii) 7 ! 5! = 5040 120 = 4920.

Example 6 Compute (i) 7!5! (ii) ( )

12!10! (2!)

Solution (i) We have 7!5! =

7 6 5!5!

= 7 6 = 42

and (ii) ( ) ( )12!

10! 2! =

( )( ) ( )

12 11 10!10! 2

= 6 11 = 66.

Example 7 Evaluate ( )!

! !n

r n r , when n = 5, r = 2.

Solution We have to evaluate (

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