2. <ul><li>Dr. R. Venkatamuni Reddy Associate Professor </li></ul><ul><li>Contact: 09632326277, 080-30938181 [email_address] [email_address] </li></ul>Basic Quantitative Techniques - RVMReddy - ABS July 14, 2010
3. Permutations and Combinations July 14, 2010 Basic Quantitative Techniques - RVMReddy - ABS
4. Permutations <ul><li>Permutations refers to the different ways in which a number of a number of objects can be arranged in a different order </li></ul><ul><li>Example: Suppose there are two things x and y, they can be arranged in to two different ways i.e,. xy and yx . These two arrangements is called permutation </li></ul><ul><li>Similarly x, y and z </li></ul><ul><li>xyz, xzy, yxz, yzx, zxy, zyx is 6 arrange permutation </li></ul><ul><li>(if we want to have two things only from x,y,z then xy,xz,yz,yx,zx,yz only in this case) </li></ul>July 14, 2010 Basic Quantitative Techniques - RVMReddy - ABS
5. <ul><li>“ The word permutation thus refers to the arrangements which can be made by taking some or all of a number of things” </li></ul><ul><li>Formulae 1: Finding the number of permutations of ‘n’ dissimilar things taken ‘r’ at a time </li></ul><ul><li>n=number of different things given, r=number of different things taken at a time out of different things given </li></ul>Permutations July 14, 2010 Basic Quantitative Techniques - RVMReddy - ABS
6. <ul><li>Example 1: There are six boxes and three balls. In how many ways can these three balls be discretely put into these six boxes. </li></ul><ul><li>Solution: </li></ul><ul><li>Example 2: How many four-letter words can be made using the letters of the word ‘BANGALORE’ and ‘ALLIANCE’ </li></ul><ul><li>Solution: n=9, r=4 and n=8, r=4 </li></ul>Permutations July 14, 2010 Basic Quantitative Techniques - RVMReddy - ABS
7. <ul><li>Example 3: How many arrangements are possible of the letters of the words ‘JAIPUR’, ‘BANGALORE’ and ‘ALLIANCE’ </li></ul><ul><li>Hint: n=6, r=6 and n=9, r=9 and n=8,r=8 </li></ul>Permutations July 14, 2010 Basic Quantitative Techniques - RVMReddy - ABS
8. <ul><li>Formulae 2: Finding the number of permutations of ‘n’ things taken ‘r’ at a time, given that each of the elements can be repeated once, twice….up to ‘r’ times </li></ul><ul><li>Or </li></ul><ul><li>‘ n’ things taken all at a time of which ‘p’ are alike, ‘q’ others are alike and ‘r’ others alike </li></ul><ul><li>Example 1: How many permutations are possible of the letters of the word PROBABILITY when taken all at a time? </li></ul>Permutations July 14, 2010 Basic Quantitative Techniques - RVMReddy - ABS
9. <ul><li>Solution: n=11, p=2 ( as letter B is occurring twice in the given word) , and q=2 ( as letter I is occurring twice in the given word) </li></ul><ul><li>And all other letters in the given word are different. The required number of permutations is (r is not valid in this) </li></ul><ul><li>Example 2: You are given a word “MANAGEMENT” and asked to compute the number of permutations that you can form taking all the letters from this word? </li></ul>Permutations July 14, 2010 Basic Quantitative Techniques - RVMReddy - ABS
11. Permutation formula proof <ul><li>There are n ways to choose the first element </li></ul><ul><ul><li>n -1 ways to choose the second </li></ul></ul><ul><ul><li>n -2 ways to choose the third </li></ul></ul><ul><ul><li>… </li></ul></ul><ul><ul><li>n - r +1 ways to choose the r th element </li></ul></ul><ul><li>By the product rule, that gives us: </li></ul><ul><li>P ( n , r ) = n ( n -1)( n -2)…( n - r +1) </li></ul>July 14, 2010 Basic Quantitative Techniques - RVMReddy - ABS
12. <ul><li>Combinations refers to the number of arrangements which can be made from a group of things irrespective of the order </li></ul><ul><li>Combinations differ from permutations in that one combination such as xyz may be stated in the form of several permutations just by rearranging the orders as : xyz, xzy, yxz, yzx, zxy, zyx </li></ul><ul><li>Note: All of these are one combination but they are six permutations </li></ul><ul><li>IMP Note: The number of permutations is always greater than the number of combinations in any given situation since a combination of n different things can be generate n factorial permutations </li></ul>Combinations July 14, 2010 Basic Quantitative Techniques - RVMReddy - ABS
13. <ul><li>Formulae 1: The number of r -combinations of a set with n elements, where n is non-negative and 0≤ r ≤ n is: </li></ul><ul><li>n= number of different things given </li></ul><ul><li>r= number of different things taken at a time out of different things given </li></ul>Combinations July 14, 2010 Basic Quantitative Techniques - RVMReddy - ABS
14. <ul><li>Example 1: in how many ways can four persons be chosen out of seven? </li></ul><ul><li>n=7, r=4 </li></ul><ul><li>Example 2: Find the number of combinations of 50 things taking 46 at a time. ANS: 230300 </li></ul>Combinations July 14, 2010 Basic Quantitative Techniques - RVMReddy - ABS
15. <ul><li>Formulae 2: The number of ways in which x+y+z things can be divided into three groups contain x, y, and z things respectively is </li></ul><ul><li>Example: In how many ways can 10 books be put to three shelves which can contain 2, 3 and 5 books respectively? </li></ul>Combinations July 14, 2010 Basic Quantitative Techniques - RVMReddy - ABS
18. Combinations <ul><li>How many different poker hands are there (5 cards)? </li></ul><ul><li>How many different (initial) blackjack hands are there? </li></ul>July 14, 2010 Basic Quantitative Techniques - RVMReddy - ABS
19. Combination formula proof <ul><li>Let C (52,5) be the number of ways to generate unordered poker hands </li></ul><ul><li>The number of ordered poker hands is P (52,5) = 311,875,200 </li></ul><ul><li>The number of ways to order a single poker hand is P (5,5) = 5! = 120 </li></ul><ul><li>The total number of unordered poker hands is the total number of ordered hands divided by the number of ways to order each hand </li></ul><ul><li>Thus, C (52,5) = P (52,5)/ P (5,5) </li></ul>July 14, 2010 Basic Quantitative Techniques - RVMReddy - ABS
20. Combination formula proof <ul><li>Let C ( n , r ) be the number of ways to generate unordered combinations </li></ul><ul><li>The number of ordered combinations (i.e. r -permutations) is P ( n , r ) </li></ul><ul><li>The number of ways to order a single one of those r -permutations P ( r,r ) </li></ul><ul><li>The total number of unordered combinations is the total number of ordered combinations (i.e. r -permutations) divided by the number of ways to order each combination </li></ul><ul><li>Thus, C ( n,r ) = P ( n,r )/ P ( r,r ) </li></ul>July 14, 2010 Basic Quantitative Techniques - RVMReddy - ABS
21. Combination formula proof <ul><li>Note that the textbook explains it slightly differently, but it is same proof </li></ul>July 14, 2010 Basic Quantitative Techniques - RVMReddy - ABS
22. <ul><li>Let n and r be non-negative integers with r ≤ n . Then C ( n , r ) = C ( n , n-r ) </li></ul><ul><li>Proof: </li></ul>Combination formula proof July 14, 2010 Basic Quantitative Techniques - RVMReddy - ABS
23. Binomial Coefficients <ul><li>The expression x + y is a binomial expression as it is the sum of two terms. </li></ul><ul><li>The expression (x + y) n is called a binomial expression of order n . </li></ul>Basic Quantitative Techniques - RVMReddy - ABS
26. Binomial Coefficients <ul><li>Pascal’s Triangle </li></ul><ul><ul><li>The number C(n , r) can be obtained by constructing a triangular array. </li></ul></ul><ul><ul><li>The row 0, i.e., the first row of the triangle, contains the single entry 1 . The row 1, i.e., the second row, contains a pair of entries each equal to 1 . </li></ul></ul><ul><ul><li>Calculate the n t h row of the triangle from the preceding row by the following rules: </li></ul></ul>Basic Quantitative Techniques - RVMReddy - ABS
29. Binomial Coefficients <ul><li>The technique known as divide and conquer can be used to compute C(n , r ). </li></ul><ul><li>In the divide-and-conquer technique, a problem is divided into a fixed number, say k , of smaller problems of the same kind. </li></ul><ul><li>Typically, k = 2 . Each of the smaller problems is then divided into k smaller problems of the same kind, and so on, until the smaller problem is reduced to a case in which the solution is easily obtained. </li></ul><ul><li>The solutions of the smaller problems are then put together to obtain the solution of the original problem. </li></ul>July 14, 2010 Basic Quantitative Techniques - RVMReddy - ABS
30. Thank You Basic Quantitative Techniques - RVMReddy - ABS July 14, 2010