*Karnataka Class 11 Commerce Maths Permutations and Combinations*

*Karnataka Class 11 Commerce Maths Permutations and Combinations*

*Karnataka Class 11 Commerce Maths Permutations and Combinations : The aim of education is not the acquisition of information, although important, or acquisition of technical skills, though essential in modern society, but the development of that bent of mind, attitude of reason, that spirit of democracy which will make us responsible citizens.*

*Karnataka Pre-University board is a government body which organizes the higher secondary examination in the state. The board functions under the Department of Primary & Secondary Education. The board has a total of 1202 government pre-university colleges, 165 unaided Pre-University colleges and about 13 Corporation pre-university colleges.*

*Each year at Karnataka board of pre-university education, about 10 lakh students enroll in the 2 year Pre-University courses. The courses offered by the Pre-University board Karnataka are Humanities (Arts), Science & Commerce. There are 23 subjects, 11 languages and 50 combinations in the Pre-University curriculum. Further, for the academic year 2010-11, the Karnataka Pr-University examination board has 4,43,185 students in Humanities in I & II PUC, 2,47,421 students in Science in I & II PUC & 2,77,189 students in commerce in I & II PUC.*

*Karnataka Class 11 Commerce Maths Permutations and Combinations*

**Syllabus – Topics**

*Permutations and Combinations**Fundamental principle of counting.**Factorial n**Permutations : Definition, examples , derivation of formulae*^{n}Pr,*Permutation when all the objects are not distinct , problems.**Combinations: Definition, examples**Proving*^{n}Cr=^{n}Pr r!,^{n}Cr =^{n}Cn-r ;^{n}Cr +^{n}Cr-1 =n+^{1}Cr*Problems based on above formulae.*

*Karnataka Class 11 Commerce Maths Permutations and Combinations*

*PERMUTATIONS *

*This topic deals with the new Mathematical idea of counting without doing actual counting. That is without listing out particular cases it is possible to assess the number of cases under certain given conditions. Permutations refer to different arrangement of things from a given lot taken one or more at a time. *

*For example, Permutations made out of a set of three elements {a,b,c} *

*(i) One at a time: {a}, {b}, {c} …… 3 ways *

*(ii) Two at a time: {a,b}, {b,a},{b,c}, {c,b}, {a,c}, {c,a} …… 6 ways *

*(iii) Three at a time: {a,b,c}, {a,c,b}, {b,c,a}, {b,a,c}, {c,a,b}, {c,b,a} ……6 ways *

*Karnataka Class 11 Commerce Maths Permutations and Combinations*

*Fundamental rules of counting *

*There are two fundamental rules of counting based on the simple principles of multiplication and addition, the former when events occur independently one after another and latter when either of the events can occur simultaneously. Some times we have to combine the two depending on the nature of the problem. *

*Karnataka Class 11 Commerce Maths Permutations and Combinations*

*Fundamental principle of counting *

*Let us consider an example from our day-to-day life. Sekar was allotted a roll number for his examination. But he forgot his number. What all he remembered was that it was a two digit odd number. *

*The possible numbers are listed as follows:*

11 | 21 | 31 | 41 | 51 | 61 | 71 | 81 | 91 |

13 | 23 | 33 | 43 | 53 | 63 | 73 | 83 | 93 |

15 | 25 | 35 | 45 | 55 | 65 | 75 | 85 | 95 |

17 | 27 | 37 | 47 | 57 | 67 | 77 | 87 | 97 |

19 | 29 | 39 | 49 | 59 | 69 | 79 | 89 | 99 |

* **So the total number of possible two digit odd numbers = 9×5 = 45 *

*Let us see whether there is any other method to find the total number of two digit odd numbers. Now the digit in the unit place can be any one of the five digits 1,3,5,7,9. This is because our number is an odd number. The digit in the ten’s place can be any one of the nine digits 1,2,3,4,5,6,7,8,9. Thus there are five ways to fill up the unit place and nine ways to fill up the ten’s place. So the total number of two digit odd numbers = 9×5 = 45. *

*Karnataka Class 11 Commerce Maths Permutations and Combinations*

*This example illustrates the following principle. *

*(i) Multiplication principle *

*If one operation can be performed in “m” different ways and another operation can be performed in “n” different ways then the two operations together can be performed in ‘m x n’ different ways. This principle is known as multiplication principle of counting.*

*(ii) Addition Principle *

*If one operation can be performed in m ways and another operation can be performed in n ways, then any one of the two operations can be performed in m+n ways. This principle is known as addition principle of counting. Further consider the set {a,b,c,d}.*

*From the above set we have to select two elements and we have to arrange them as follows*

*Karnataka Class 11 Commerce Maths Permutations and Combinations*

*Karnataka Class 11 Commerce Maths Permutations and Combinations*

*Karnataka Class 11 Commerce Maths Permutations and Combinations*

*Karnataka Class 11 Commerce Maths Permutations and Combinations*

*The possible arrangements are (a,b), (a,c), (a,d) (b,a), (b,c), (b,d) (c,a), (c,b), (c,d) (d,a), (d,b), (d,c).*

* The total number of arrangements are 4 × 3 = 12. *

*In the above arrangement, the pair (a,b) is different from the pair (b,a) and so on. *

*There are 12 possible ways of arranging the letters a,b,c,d taking two at a time. i.e Selecting and arranging ‘2’ from ‘4’ can be done in 12 ways. *

*In otherwords number of permutations of ‘four’ things taken ‘two’ at a time is 4 × 3 = 12.*

*In general ^{n}P_{r }denotes the number of permutations of ‘n’ things taken ‘r’ at a time. [‘n’ and ‘r’ are positive integers and r ≤ n].*

*Karnataka Class 11 Commerce Maths Permutations and Combinations*

*To find the value of ^{n}P_{r } : *

* ^{n}P_{r } means selecting and arranging ‘r’ things from ‘n’ things which is the same as filling ‘r’ places using ‘n’ things which can be done as follows. *

*The first place can be filled by using anyone of ‘n’ things in ‘n’ ways. *

*The second place can be filled by using any one of the remaining (n – 1) things in (n – 1) ways. So the first and the second places together can be filled in n(n – 1) ways. *

*The third place can be filled in (n – 2) ways by using the remaining (n – 2) things. So the first, second and the third places together can be filled in n (n – 1) (n – 2) ways. In general ‘r’ places can be filled in n(n – 1)(n – 2)….[n – (r – 1)] ways. So npr = n(n – 1) (n – 2)…(n – r + 1). To simplify the above formula, we are going to introduce factorial notation.*

*Karnataka Class 11 Commerce Maths Permutations and Combinations*

**Factorial notation: **

*The product of first ‘n’ natural numbers is called n- factorial denoted by n ! or n . *

*For example: *

*5! = 5 × 4 × 3 × 2 × 1 *

*4! = 4 × 3 × 2 × 1 *

*∴ 5! = 5 × 4! *

*5! = 5 × 4 × 3! *

*In general, n! = n (n – 1) (n – 2)…3.2.1 *

*∴ n! = n{(n – 1)!} *

*= n (n – 1)(n – 2)! and so on *

*We have ^{n}P_{r }= n (n – 1)(n – 2)…………(n – r + 1) *

*= n (n – 1)(n – 2)…………(n – r + 1) (n – r)! = n! /(n – r)! *

*{multiplying and dividing by (n – r)!} *

*∴ ^{n}P_{r} = n! /(n – r)! .*

*Karnataka Class 11 Commerce Maths Permutations and Combinations*

*Observation : *

*(i) 0! = 1 *

*(ii) ^{n}P_{0} = n! /(n – 0)! = n!/n! = 1*

*(iii) ^{n}P_{1} = n! /(n – 1)! = n(n – 1)! /(n – 1)! = n*

*(iv) ^{n}P_{n}= n! /(n – n)! = n! /0! = 0 *

*(ie. Selecting and arranging ‘n’ things from ‘n’ things can be done in n! ways). *

*(i.e ‘n’ things can be arranged among themselves in n! ways).*

*Karnataka Class 11 Commerce Maths Permutations and Combinations*

**Permutations of repeated things:**

*If there are ‘n’ things of which ‘m’ are of one kind and the remaining (n-m) are of another kind, then the total number of distinct permutations of ‘n’ things = n! /m! (n – m)!*

*If there are m _{1} things of first kind, m_{2} things of second kind and m_{r} things of rth kind such that m_{1} + m_{2} + …..+ m_{r} = n then the total number of permutations of ‘n’ things = n! /m_{1}! m_{2}*

*!………..*

*m*

_{r}*!.*

*Karnataka Class 11 Commerce Maths Permutations and Combinations*

*Circular Permutations: *

*We have seen permutations of ‘n’ things in a row. Now we consider the permutations of ‘n’ things in a circle. Consider four letters A,B,C,D. The four letters can be arranged in a row in 4! ways. Of the 4! arrangements, the arrangement ABCD, BCDA, CDAB, DABC are the same when represented along a circle.*

*Karnataka Class 11 Commerce Maths Permutations and Combinations*

*So the number of permutations of ‘4’ things along a circle is 4 ! /4 = 3! In general, n things can be arranged among themselves in a circle in (n – 1)! ways*

*Karnataka Class 11 Commerce Maths Permutations and Combinations*

*COMBINATIONS*

*A combination formula is a way of selecting items from a collection, such that the order of selection does not matter. The combination involves selection of objects or things out of larger group where order doesn’t matter.*

*The combination formula will find the number of possible combinations that can be obtained by taking a sub-set of items from a larger set. It shows how many different possible sub-sets can be made from the larger set.*

*The combination formula show the number of ways a sample of “r” elements can be obtained from a larger set of “n” distinguishable objects.*

*Combination are selections ie. it involves only the selection of the required number of things out of the total number of things. Thus in combination order does not matter. *

*For example, consider a set of three elements {a,b,c} and combination made out of the set with *

*i) One at a time: {a}, {b}, {c} *

*ii) Two at a time: {a,b}, {b,c}, {c,a} *

*iii) Three at a time: {a,b,c}. *

*The number of combinations of n things taken r, (r ≤ n) is denoted by ^{n}C_{r} or ( n /r).*

*Karnataka Class 11 Commerce Maths Permutations and Combinations*

*To derive the formula for ^{n}C_{r} : *

*Number of combinations of ‘n’ things taken ‘r’ at a time = ^{n}C_{r} *

*Number of permutations of ‘n’ things taken ‘r’ at a time = ^{n}P_{r}*

*Number of ways ‘r’ things can be arranged among themselves = r! *

*Each combination having r things gives rise to r! permutations.*

*∴ ^{n}P_{r = }*

*(*)^{n}C_{r}*r!**⇒ n!/(n-r)! = ( ^{n}C_{r}) r!*

**^{∴ n}C_{r}** =

*n!/r! (n-r)!**Karnataka Class 11 Commerce Maths Permutations and Combinations*

**Observation:**

**(i) ^{n}C_{0}** =

**n!/0! (n-0)! =**

**n!/n! = 1****(ii)** **^{n}C_{n}** =

*n!/n! (n-n)! =**n!/n! = 1***(iii)** **^{n}C_{r}** =

^{n}C_{n-r}**(iv)** if **^{n}C_{x}** =

^{n}C_{y then x = y or x + y = n}**(v)** **^{n}C_{r}** =

^{n}P_{r}**/ r!.***Karnataka Class 11 Commerce Maths Permutations and Combinations*

**Pascal’s Triangle **

*For n = 0, 1, 2, 3, 4, 5 … the details can be arranged in the form of a triangle known as Pascal’s triangle.*

*Karnataka Class 11 Commerce Maths Permutations and Combinations*

*Substituting the values we get*

*Karnataka Class 11 Commerce Maths Permutations and Combinations*

*The conclusion arrived at from this triangle named after the French Mathematician Pascal is as follows. The value of any entry in any row is equal to sum of the values of the two entries in the preceding row on either side of it. Hence we get the result.*

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