Quantitative MethodsSession – 29.08.13Permutation & Combination

Pranjoy Arup Das

PERMUTATION

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) ABC2) ACB3) BCA4) BAC5) CAB6) 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.

• 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) AB2) BC3) CA4) BA5) CB6) 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.

• Eg3. Find the permutation of the letters A, B, C & D taking three letters at a time. 1. ABC2. ABD3. ACD4. ACB5. ADB6. ADC7. BCD8. BCA9. BDA10. BDC11. BAC12. BAD13. CDA14. CDB15. CBA16. CBD17. CAB18. CAD19. DAB20. DAC21. DBC22. DBA23. DCB24. DCA Permutation of the 4 letters A,B,C,D taking 3 at a time is 24.

• 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 nPr or P (n, r), read as ‘Permutation of n things taking r things at a time’.nPr or 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

• 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 = 4r = 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

• 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 = = ___________

• 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 =

66P6

• 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 = _____________

5P335

COMBINATION

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) AB2) AC3) 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.

• 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 timec) taking all 4 letters at a time. Solution: a) 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.

• 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 nCr or C (n, r) is given by : nCr or 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

• 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

nCr or C (n, r)= 32C4 = = == = So the required number of combination s =______________

• 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 nCr or C (n, r)= 6C2 = 6 ! / (6-2) ! * 2 ! = 6 * 5 / 2 = 15• So the required number of combination s = 15

• 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 nCr or 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

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

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= (1st Combn of Part A * 1st Combn of Part B ) + ( 2nd combination of A * 2nd combn 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 woman2nd 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 = 5C1and 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

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 women2nd combination - Selecting 6 men and 6 women. 3rd combination – selecting 5 men and 7 women4th combination – selecting 4 men and 8 women5th 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 = 8C6and 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

1) In how many ways can seven pictures be hug from five nails on the wall? 2) Three men have 4 coats, 5 trousers and 6 caps. In how many ways can

they wear them? 3) How many four letter words can be formed from the word LOGARITHMS,

without repeating any letter? 4) 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?

5) 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?

6) 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?

7) 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?

8) 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?

9) 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.

END OF CHAPTER 10