Quantitative Methods Session – 29.08.13 Permutation & Combination Pranjoy Arup Das

1. Quantitative MethodsSession – 29.08.13Permutation & CombinationPranjoy Arup Das

2. PERMUTATION

3. Permutation : • The number of ways in which a set or a group of things can be arranged in different ways is termed as permutation. • For Eg1. Suppose we have a set of the three letters A,B and C. These 3 letters can be arranged in the following ways: 1) ABC 2) ACB 3) BCA 4) BAC 5) CAB 6) CBA • So there are 6 ways in which A,B & C can be arranged. • In other words, the permutation of the three letters A,B & C is 6. • It will be more appropriate to say that the permutation of the 3 letters A, B & C taking all three at a time is 6.

4. Eg2. Suppose we have a set of the three letters A,B and C. In how many ways can the three be arranged taking two letters at a time? • The 3 letters can be arranged in the following ways taking two at a time: 1) AB 2) BC 3) CA 4) BA 5) CB 6) AC • So there are 6 ways in which the letters A,B & C can be arranged taking two at a time. • In other words, the permutation of the 3 letters A, B & C taking 2 letters at a time is 6. • In permutation, alongwith arrangement, the order of placement of the things is also equally important.

5. Eg3. Find the permutation of the letters A, B, C & D taking three letters at a time. • ABC • ABD • ACD • ACB • ADB • ADC • BCD • BCA • BDA • BDC • BAC • BAD • CDA • CDB • CBA • CBD • CAB • CAD • DAB • DAC • DBC • DBA • DCB • DCA Permutation of the 4 letters A,B,C,D taking 3 at a time is 24.

6. THEOREM OF PERMUTATION: If there are ‘n’ number of different things to be arranged by taking ‘r’ things at a time, (where r ≥ 1 and r ≤ n), then the permutation, denoted by nPror P (n, r), read as ‘Permutation of n things taking r things at a time’. nPror P (n, r)= POINTS TO NOTE : n ! = n * (n-1) * (n-2) * (n-3) *………………………………3 * 2 * 1 Eg. 12 ! = 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 479001600 (n-r) ! = (n-r) * (n-r-1) * (n-r-2) * (n-r-3)* ……………3 * 2 * 1 IMPORTANT POINT 1! = 1 0 ! = 1

7. Eg3. Using permutation theorem, find the permutation of the letters A, B, C & D taking three letters at a time. • Solution: There are 4 letters to be arranged taking 3 letters at a time. n = 4 r = 3 nPr= 4P3 = = = = • So the permutation of the 4 letters A,B,C & D taking 3 letters at a time is 24. • Eg. 4 Find the values of 5P3 , 5P4 and 5P5. 5P3 = 5 ! / (5-3)! = 5 ! / 2 ! = (5 * 4 * 3 * 2 * 1) / (2 * 1) = _______ 24

8. Eg5. You have 3 rings. In how many different ways can the 3 rings be worn on four fingers. (One ring on each finger ). • Solution: > If there are 3 rings then, at a time, they can be worn on 3 of the 4 fingers. > That means out of 4 fingers, we have to take 3 fingers at a time. > So, here n = 4 & r = 3 > So the permutation of wearing the 3 rings on 4 fingers taking 3 fingers at a time : 4P3 = = ___________

9. Eg5. Seven athletes are participating in a race. There are three prizes, 1st , 2nd & 3rd. In how many ways can the first three prizes be won? • Solution: > If there are 7athletes and there are 3 prizes, that means at one time only 3 athletes can be given the 3 prizes. > So, here n = 7 and r = 3 > And the permutation of the 7 athletes winning the 3 prizes taking 3 athletes at a time is : 7P3 = 7! / (7-3)! = ________ • Eg 6 In how many ways can 6 persons stand together in a queue? • Solution: If the 6 persons are to stand in queue together, then at any point of time, all 6 of them will have to be in the queue. > So here, n = 6 and r = And permutation = 6 6P6

10. Eg6. x is a number between 100 and 1000 such that 100 <x<1000. How many values of x can be formed from the digits 1,2,3,4 & 5? No digit can be repeated. ( numbers like 111, 221, 433, 545 etc. not allowed) Solution : > It is obvious that any number between 100 and 1000 will be a three digit number. > This means that out of our total of 5 digits, we have to use 3 digits at a time. So here, n = , r = And the permutation = = ______________ • Eg 7, How many words can be formed using all the letters, at one time, of the word EQUATION. The words may or may not have any meaning. • Solution: The word equation has 8 letters from which we have to form words using all 8 letters, i.e., 8 letter words. • So here, n = _______, r = ___________, and permutation = _____________ 5 3 5P3

11. COMBINATION

12. Combination : • The number of things that can be selected at a time from a larger group of things is termed as combination. • For Eg1. Suppose we have a set of the three letters A,B and C. We want to form a combination of these 3 taking 2 at a time. These 3 can be selected in the following ways: 1) AB 2) AC 3) BC • So there are 3 possible combinations of A,B & C taking 2 at a time. • Unlike permutation, in combination the arrangement or order of placement of the things has no importance. • Eg. The permutation of A, B & C taking 2 at a time are AB,BA,AC,CA, BC, CB, i.e., 6. • In combination, AB & BA mean the same, AC & CA mean the same etc.

13. Eg2. Suppose we have a set of the four letters A,B, C & D. We want to form a combination of these 4 letters: a) taking 2 letters at a time. b) taking 3 letters at a time c) taking all 4 letters at a time. Solution: • These 4 letters can be selected in the following ways: AB, AC, AD, BC, BD, CD • So there are 6 possible combinations of A,B, C &D taking 2 at a time. b) These 4 letters can be selected in the following ways : ABC, ABD, ACD, BCD • So there are 4 possible combinations of A,B,C & D taking 3 at a time. c) The combinations are : ABCD • So there is 1 possible combination of A,B,C & D taking all 4 at a time.

14. THEOREM OF COMBINATION: If there are ‘n’ number of different things to be selected by taking ‘r’ things at a time, where r ≥ 0 and r ≤ n, then the combination, denoted by nCror C (n, r) is given by : nCror C (n, r)= POINTS TO NOTE : • n ! = n * (n-1) * (n-2) * (n-3) *………………………………3 * 2 * 1 Eg. 12 ! = 12 * 11 * 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 479001600 • (n-r) ! = (n-r) * (n-r-1) * (n-r-2) * (n-r-3)* ……………3 * 2 * 1 • r ! = r * (r-1) * (r-2) * (r-3)* ……………3 * 2 * 1

15. Eg. 3) From a class of 32 students, 4 students are to be chosen for a competition. In how many ways can 4 students be selected? • Solution: There are 32 students from which 4 students are to be selected. This means that out of 32 students, at a time 4 students can be selected. We need to find the no. of possible combinations: Where, n = 32, r = 4 nCror C (n, r)= 32C4 = = = = = So the required number of combination s =______________

16. Eg. 4) Three gentlemen and three ladies are candidates to be selected through voting for two vacancies. A voter has to vote for the two vacancies separately. In how many ways can the voter cast his vote? • Solution: • There are a total of 6 candidates ( 3 male + 3 female) from which 2 candidates are to be selected. • This means that out of 6 candidates, at a time 2 candidates can be selected. • We need to find the no. of combinations possible : Where , n = 6, r = 2 nCror C (n, r)= 6C2 = 6 ! / (6-2) ! * 2 ! = 6 * 5 / 2 = 15 • So the required number of combination s = 15

17. Eg. 5) A question paper has two parts, Part A & Part B each containing 10 questions. If students have to answer any 8 questions from Part A and any 5 questions from Part B, in how many ways can the questions be selected? • Solution: > There are a total of 10 questions in Part A from which 8 questions are to be selected. > This means that out of 10 questions of part A, at a time 8 questions can be selected. > So in this case n = 10, r = 8, And nCror C (n, r)= 10C8 > Again in Part B there are 10 questions from which 5 questions are to be answered. > So in the case of Part B, n = 10, r = 5 and nCr= 10C5 > Therefore, the total no. of combinations = 10C8 *10C5

18. POINT TO NOTE: • If a group has two parts A & B, then: Total Combination of the group= Combination of Part A * Combination of Part B Eg 5) In how many ways can 6 men and 5 women form a committee of 5 members comprising of 3 men and 2 women? Solution: • Out of 11 persons, we need to select 5 members to form a committee. • But out of these 11 persons, 6 are men and 5 are women and the new committee has to have 3 men and 2 women. • Therefore, Combination of 6 men taking 3 men at a time = 6C3 • and the combination of 5 women taking 2 women at a time = 5C2 • Therefore out of the 11 persons, a 5 member committee consisting of 3 men and 2 women can be formed in : 6C3 *5C2 = ___________Combinations

19. POINT TO NOTE: • If a group has two parts A & B and two combinations of A & B are possible then: • Total Combination of the group= (1stCombn of Part A * 1stCombn of Part B ) + ( 2nd combination of A * 2ndcombn of B) Eg 6) In how many ways can 5 men and 2 women form a committee of 3 members having at least 1 woman? • The 3 member committee has to have at least 1 woman. In such a case, the following combinations are possible: 1st Combination = Selecting 2 men and 1 woman 2nd combination =Selecting 1 man and 2 women. • Therefore, 1st Combination of 5 men taking 2 men at a time = 5C2 and the 1st combination of 2 women taking 1 woman at a time = 2C1 • 2nd combination of 5 men taking 1 man at a time = 5C1 and the 2nd combination of 2 women taking 2 women at a time = 2C2 • Therefore out of the 7 persons, a 3 member committee consisting of at least 1 woman can be formed in : (5C2 * 2C1 )+ (5C1 * 2C2 ) = __________Combns

20. Eg 6) In how many ways can 8 men and 9 women form a committee of 12 members comprising of at least 5 women? The 12 member committee has to have at least 5 women. In such a case, the following combinations are possible: 1st Combination - Selecting 7 men and 5 women 2nd combination - Selecting 6 men and 6 women. 3rd combination – selecting 5 men and 7 women 4th combination – selecting 4 men and 8 women 5th combination – selecting 3 men and 9 women • Therefore, 1st Combination of 8 men taking 7 men at a time = 8C7 and the 1st combination of 9 women taking 5 women at a time =9C5 • 2nd combination of 8 men taking 6 men at a time = 8C6 and the 2nd combination of 9 women taking 6 women at a time = 9C6 • 3rd combo of men= 8C5, 3rd combo of women = 9C7 • 4th combo of men =8C4 , 4th combo of women =9C8 • 5th combo of men = 8C3 , 5th combo of women = 9C9 • Therefore out of the 17persons, a 12 member committee consisting of at least 5 women can be formed in : (8C7 * 9C5) + (8C6 * 9C6) + (8C5 * 9C7) + (8C4 * 9C8) + (8C3 + 9C9

21. In how many ways can seven pictures be hug from five nails on the wall? • Three men have 4 coats, 5 trousers and 6 caps. In how many ways can they wear them? • How many four letter words can be formed from the word LOGARITHMS, without repeating any letter? • There are six periods in each working day of a school. In how many ways can 5 subjects be arranged such that each subject is allowed at least one subject? • In how many ways can 6 boys and 5 girls be arranged for a group photo if the girls are to sit on chairs in a row and the boys are to stand in a row behind them? • Out of 5 men and 2 women, a committee of 3 is to be formed. In how many ways can it be formed if at least one woman is to be included? • In an examination a student has to answer 4 questions out of 5 questions. Question nos. 1 & 2 and compulsory. In how many ways can a student select his questions? • In a room there are three sofas. One is a three-seater sofa, another a four-seater and the third a five –seater. There are 12 guests. In how many ways can the 12 guests be selected to sit on each sofa? • From a class of 25 students, 10 students are to be selected for an excursion. 3 students have already been selected. In how many ways can the excursion party be formed? There is a possibility that the 3 students may not join the party.

22. END OF CHAPTER 10