Combinations and Permutations Whats the Difference? In English we use the word "combination" loosely, without thinking if the order of things is important. In other words: "My fruit salad is a combination of apples, grapes and bananas" We dont care what order the fruits are in, they could also be "bananas, grapes and apples" or "grapes, apples and bananas", its the same fruit salad. "The combination to the safe was 472". Now we do care about the order. "724" would not work, nor would "247". It has to be exactly 4-7-2. So, in Mathematics we use more precise language: If the order doesnt matter, it is a Combination. If the order does matter it is a Permutation. A Permutation is an ordered Combination.Permutation : Permutation means arrangement of things. Theword arrangement is used, if the order of things is considered.Combination: Combination means selection of things. The wordselection is used, when the order of things has no importance.Definition:Permutation: An arrangement is called a Permutation. It is therearrangement of objects or symbols into distinguishable sequences.When we set things in order, we say we have made an arrangement.When we change the order, we say we have changed the
arrangement. So each of the arrangement that can be made bytaking some or all of a number of things is known as Permutation.Combination: A Combination is a selection of some or all of a number ofdifferent objects. It is an un-ordered collection of unique sizes.In apermutation the order of occurence of the objects or thearrangement is important but in combination the order of occurenceof the objects is not important.Formula:Permutation = nPr = n! / (n-r)!Combination = nCr = nPr / r!where, n, r are non negative integers and 0≤ r≤n. r is the size of each permutation. n is the size of the set from which elements are permuted. ! is the factorial operator.Example: Suppose we have to form a number of consistingof three digits using the digits 1,2,3,4, To form this number thedigits have to be arranged. Different numbers will get formeddepending upon the order in which we arrange the digits. Thisis an example of Permutation.Now suppose that we have to make a team of 11 players out of20 players, This is an example of combination, because theorder of players in the team will not result in a change in theteam. No matter in which order we list out the players the teamwill remain the same! For a different team to be formed at leastone player will have to be changed.Now let us look at two fundamental principles of counting:
Addition rule : If an experiment can be performed in „n‟ ways, &another experiment can be performed in „m‟ ways then either ofthe two experiments can be performed in (m+n) ways. This rulecan be extended to any finite number of experiments.Example: Suppose there are 3 doors in a room, 2 on oneside and 1 on other side. A man wants to go out from the room.Obviously he has „3‟ options for it. He can come out by door „A‟or door „B‟ or door ‟C‟.Multiplication Rule : If a work can be done in m ways, anotherwork can be done in „n‟ ways, then both of the operations canbe performed in m x n ways. It can be extended to any finitenumber of operations.Example.: Suppose a man wants to cross-out a room, whichhas 2 doors on one side and 1 door on other site. He has 2 x1 = 2 ways for it.Factorial n : The product of first „n‟ natural numbers is denotedby n!. n! = n(n-1) (n-2) ………………..3.2.1. Ex. 5! = 5 x 4 x 3 x 2 x 1 =120 Note 0! = 1
Proof n! =n, (n-1)! Or (n-1)! = [n x (n-1)!]/n = n! /n Putting n = 1, we have O! = 1!/1 = 1Permutation Number of permutations of „n‟ different things taken „r‟ at a time is given by:- n Pr = n!/(n-r)!Proof: Say we have „n‟ different things a1, a2……, an.Clearly the first place can be filled up in „n‟ ways. Number of things left after filling-up the first place = n-1So the second-place can be filled-up in (n-1) ways. Now number of things left after filling-up the first and second places = n - 2Now the third place can be filled-up in (n-2) ways.Thus number of ways of filling-up first-place = nNumber of ways of filling-up second-place = n-1Number of ways of filling-up third-place = n-2Number of ways of filling-up r-th place = n – (r-1) = n-r+1By multiplication – rule of counting, total no. of ways of filling up, first, second -- rth-place together :-
n (n-1) (n-2) ------------ (n-r+1) Hence:n Pr = n (n-1)(n-2) --------------(n-r+1)= [n(n-1)(n-2)----------(n-r+1)] [(n-r)(n-r-1)-----3.2.1.] / [(n-r)(n-r-1)] ----3.2.1 n Pr = n!/(n-r)! Number of permutations of „n‟ different things taken all at a time is given by:- n Pn = n!Proof :Now we have „n‟ objects, and n-places.Number of ways of filling-up first-place = nNumber of ways of filling-up second-place = n-1Number of ways of filling-up third-place = n-2Number of ways of filling-up r-th place, i.e. last place =1Number of ways of filling-up first, second, --- n th place= n (n-1) (n-2) ------ 2.1. n Pn = n!Concept. nWe have Pr = n!/n-rPutting r = n, we have :-
n Pr = n! / (n-r) n But Pn = n!Clearly it is possible, only when n! = 1Hence it is proof that 0! = 1Note : Factorial of negative-number is not defined. Theexpression –3! has no meaning.ExamplesQ. How many different signals can be made by 5 flags from 8-flags of different colours?Ans. Number of ways taking 5 flags out of 8-flage = 8P5= 8!/(8-5)!= 8 x 7 x 6 x 5 x 4 = 6720Q. How many words can be made by using the letters of the word “SIMPLETON” taken all at a time?Ans. There are „9‟ different letters of the word “SIMPLETON”Number of Permutations taking all the letters at a time = 9P9= 9! = 362880.Number of permutations of n-thing, taken all at a time, in which „P‟ are of one type, „q‟ of them are of second-type, „r‟ of them are of third-type, and rest are all different is given by :- n!/(p! x q! xr!)
Example: In how many ways can the letters of the word “Pre- University” be arranged?13!/(2! X 2! X 2!)Number of permutations of n-things, taken „r‟ at a time when each thing can be repeated r-times is given by = nr.Proof.Number of ways of filling-up first –place = nSince repetition is allowed, so Number of ways of filling-up second-place = nNumber of ways of filling-up third-placeNumber of ways of filling-up r-th place = nHence total number of ways in which first, second ----r th, places can be filled-up= n x n x n ------------- r factors.= nrExample: John has 8 friends. In how many ways can he inviteone or more of them to dinner?Ans. John can select one or more than one of his 8 friends.=> Required number of ways = 28 – 1= 255.
(iv) Number of ways of selecting zero or more things from „n‟ identical things is given by :- n+1Example: In how many ways, can zero or more letters be selected form the letters AAAAA?Ans. Number of ways of : Selecting zero As = 1 Selecting one As = 1 Selecting two As =1 Selecting three As = 1 Selecting four As = 1 Selecting five As = 1=> Required number of ways = 6 [5+1](V) Number of ways of selecting one or more things from „p‟ identical things of one type „q‟ identical things of another type, „r‟ identical things of the third type and „n‟ different things is given by :-(p+1) (q+1) (r+1)2n – 1
Example: Find the number of different choices that can be made from 3 apples, 4 bananas and 5 mangoes, if at least one fruit is to be chosen.Ans:Number of ways of selecting apples = (3+1) = 4 ways.Number of ways of selecting bananas = (4+1) = 5 ways.Number of ways of selecting mangoes = (5+1) = 6 ways.Total number of ways of selecting fruits = 4 x 5 x 6But this includes, when no fruits i.e. zero fruits is selected=> Number of ways of selecting at least one fruit = (4x5x6) -1 = 119Note :- There was no fruit of a different type, hence here n=o=> 2n = 20=1(VI) Number of ways of selecting „r‟ things from „n‟ identical things is „1‟.Example: In how many ways 5 balls can be selected from „12‟ identical red balls?Ans. The balls are identical, total number of ways of selecting 5 balls = 1.Example: How many numbers of four digits can be formed with digits 1, 2, 3, 4 and 5?
Ans. Here n = 5 [Number of digits]And r = 4 [ Number of places to be filled-up] 5Required number is P4 = 5!/1! = 5 x 4 x 3 x 2 x 1Example: A child has 3 pocket and 4 coins. In how many ways can he put the coins in his pocket.Ans. First coin can be put in 3 ways, similarly second, third and forth coins also can be put in 3 ways.So total number of ways = 3 x 3 x 3 x 3 = 34 = 81 So, we should really call this a "Permutation Lock"! Permutations There are basically two types of permutation: 1. Repetition is Allowed: such as the lock above. It could be "333". 2. No Repetition: for example the first three people in a running race. You cant be first and second.
1. Permutations with Repetition These are the easiest to calculate.When you have n things to choose from ... you have n choices each time! When choosing r of them, the permutations are: n × n × ... (r times) (In other words, there are n possibilities for the first choice, THENthere are n possibilites for the second choice, and so on, multplying each time.) Which is easier to write down using an exponent of r: n × n × ... (r times) = nr Example: in the lock above, there are 10 numbers to choose from (0,1,..9) and you choose 3 of them: 10 × 10 × ... (3 times) = 103 = 1,000 permutations So, the formula is simply: nr where n is the number of things to choose from, and you choose r of them (Repetition allowed, order matters)
2. Permutations without RepetitionIn this case, you have to reduce the number of available choices each time. For example, what order could 16 pool balls be in? After choosing, say, number "14" you cant choose it again.So, your first choice would have 16 possibilites, and your next choice would then have 15 possibilities, then 14, 13, etc. And the total permutations would be: 16 × 15 × 14 × 13 × ... = 20,922,789,888,000But maybe you dont want to choose them all, just 3 of them, so that would be only: 16 × 15 × 14 = 3,360In other words, there are 3,360 different ways that 3 pool balls could be selected out of 16 balls. But how do we write that mathematically? Answer: we use the "factorial function" The factorial function (symbol: !) just means to multiply a series of descending natural numbers. Examples: 4! = 4 × 3 × 2 × 1 = 24 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5,040 1! = 1 Note: it is generally agreed that 0! = 1. It may seem funny that multiplying no numbers together gets you 1,
but it helps simplify a lot of equations. So, if you wanted to select all of the billiard balls the permutations would be: 16! = 20,922,789,888,000 But if you wanted to select just 3, then you have to stop themultiplying after 14. How do you do that? There is a neat trick ... you divide by 13! ... 16 × 15 × 14 × 13 × 12 ... = 16 × 15 × 14 = 3,360 13 × 12 ... Do you see? 16! / 13! = 16 × 15 × 14 The formula is written: where n is the number of things to choose from, and you choose r of them (No repetition, order matters) Examples: Our "order of 3 out of 16 pool balls example" would be: 16! 16! 20,922,789,888,000 = = = 3,360 (16-3)! 13! 6,227,020,800 (which is just the same as: 16 × 15 × 14 = 3,360)
How many ways can first and second place be awarded to 10 people? 10! 10! 3,628,800 = = = 90 (10-2)! 8! 40,320 (which is just the same as: 10 × 9 = 90) Notation Instead of writing the whole formula, people use different notations such as these: Example: P(10,2) = 90 Combinations There are also two types of combinations (remember the order does not matter now): 1. Repetition is Allowed: such as coins in your pocket (5,5,5,10,10) 2. No Repetition: such as lottery numbers (2,14,15,27,30,33)How many permutations of 3 different digits are there, chosen fromthe ten digits 0 to 9 inclusive?A84B120C
504D 720Answer :The number of permutations of 3 digits chosen from 10 is10 P3 = 10 × 9 × 8 = 720Jones is the Chairman of a committee. In how many ways can acommittee of 5 be chosen from 10 people given that Jones must beone of them?A126B252C495D 3,024 Answer: Jones is already chosen, so we need to choose another 4from 9.In choosing a committee, order doesnt matter; so we need thenumber of combinations of 4 people chosen from 9= 9 C4= 9!/(4!)(5!)= (9 × 8 × 7 × 6)/(4 × 3 × 2 × 1)= 3,024/24 = 126A password consists of four different letters of the alphabet. Howmany different possible passwords are there? (Using pen and paper can help you learn.)A426
B456,976C14,950D 358,800Answer: The number of permutations of 4 letters chosen from 26 is26 P4 = 26 × 25 × 24 × 23 = 358,800A password consists of two letters of the alphabet followed by threedigits chosen from 0 to 9. Repeats are allowed. How many differentpossible passwords are there? (Dont worry if you get it wrong ... you can learn from your mistakes.)A492,804B650,000C676,000D1,757,600Answer: The number of ways of choosing the letters = 26 × 26 = 676The number of ways of choosing the digits = 10 × 10 × 10 = 1,000So the number of possible passwords = 676 × 1,000 = 676,000An encyclopedia has eight volumes. In how many ways can the eightvolumes be replaced on the shelf? (Use pen and paper to work out the answer.)A8B
5,040C40,320D 88 Answer:Imagine there are 8 spots on the shelf. Replace the volumesone by one.The first volume to be replaced could go in any one of the eight spots.The second volume to be replaced could then go in any one of theseven remaining spots.The third volume to be replaced could then go in any one of the sixremaining spots.etcSo the total number of ways the eight volumes could be replaced= 8!=8×7×6×5×4×3×2×1= 40,320Assuming that any arrangement of letters forms a word, how manywords of any length can be formed from the letters of the wordSQUARE?(No repeating of letters) (No guessing! Be sure of your answer.)A82B720C1,956D9,331Answer: The number of one letter words = 6P1 = 6The number of two letter words = 6P2 = 6 × 5 = 30
The number of three letter words = 6P3 = 6 × 5 × 4 = 120The number of four letter words = 6P4 = 6 × 5 × 4 × 3 = 360The number of five letter words = 6P5 = 6 × 5 × 4 × 3 × 2 = 720The number of six letter words = 6P6 = 6! = 720So the total number of possible words = 6 + 30 + 120 + 360 + 720 +720 = 1,95616 teams enter a competition. They are divided up into four Pools (A,B, C and D) of four teams each.Every team plays one match against the other teams in its Pool.After the Pool matches are completed:• the winner of Pool A plays the second placed team of Pool B• the winner of Pool B plays the second placed team of Pool A• the winner of Pool C plays the second placed team of Pool D• the winner of Pool D plays the second placed team of Pool CThe winners of these four matches then play semi-finals, and thewinners of the semi-finals play in the final.How many matches are played altogether? (Using pen and paper can help you learn.)A23B31C32D63Answer:The number of matches played in each Pool = 4C2 = 4!/(2!2!)= (4 × 3)/(2 × 1) = 6
So the total number of Pool matches = 4 × 6 = 24The winners and second placed teams play a further 4 matches.Then there are 2 semi-finals and 1 finalSo the total number of matches = 24 + 4 + 2 + 1 = 31A restaurant offers 5 choices of appetizer, 10 choices of main mealand 4 choices of dessert. A customer can choose to eat just onecourse, or two different courses, or all three courses. Assuming allchoices are available, how many different possible meals does therestaurant offer?A329B310C200D19Answer; A person who eats only an appetizer has 5 choices.A person who eats only a main meal has 10 choices.A person who eats only a dessert has 4 choices.A person who eats an appetizer and a main meal has 5 × 10 = 50choices.A person who eats an appetizer and a dessert has 5 × 4 = 20 choices.A person who eats a main meal and a dessert has 10 × 4 = 40 choices.A person who eats all three courses has 5 × 10 × 4 = 200 choicesSo the total number of possible meals = 5 + 10 + 4 + 50 + 20 + 40 +200 = 329
Question:How many ways can 4 prizes be given away to 3 boys, if eachboy is eligible for all the prizes? (1)256 (2)12 (3)81 (4) None of theseCorrect Answer - (3)Solution:Any one prize can be given to any one of the 3 boys and hencethere are 3 ways of distributing each prize.Hence, the 4 prizes can be distributed in 34= 81 ways.Question:How many words of 4 consonants and 3 vowels can be madefrom 12 consonants and 4 vowels, if all the letters aredifferent? (1)16C7.7! (2)12C4.4C3.7!
(3)12C3.4C4 (4) 12C4 . 4C3Correct Answer - (2)Solution: 124 consonants out of 12 can be selected in C4 ways.3 vowels can be selected in 4C3 ways.Therefore, total number of groups each containing 4consonants and 3 vowels = 12C4 . 4C3Each group contains 7 letters, which can be arranging in 7!ways. 12Therefore required number of words = 4 . 4C3 . 7!Question:In how many ways can the letters of the word EDUCATION berearranged so that the relative position of the vowels andconsonants remain the same as in the word EDUCATION? (1)9!/4 (2)9!/(4!.5!) (3)4!.5! (4) None of theseCorrect Answer - (3)
Solution:The word EDUCATION is a 9 letter word, with none of theletters repeating.The vowels occupy 3, 5, 7th and 8th position in the word andthe remaining 5 positions are occupied by consonantsAs the relative position of the vowels and consonants in anyarrangement should remain the same as in the wordEDUCATION, the vowels can occupy only the aforementioned 4places and the consonants can occupy 1st, 2nd, 4th, 6th and 9thpositions.The 4 vowels can be arranged in the 3rd, 5th, 7th and 8thposition in 4! Ways.Similarly, the 5 consonants can be arranged in 1st, 2nd, 4th, 6thand 9th position in 5! Ways.Hence, the total number of ways = 4! . 5!.Question:There are 12 yes or no questions. How many ways can thesebe answered? (1)1024 (2)2048 (3) 4096 (4) 144Correct Answer - (3)
Solution:Each of the questions can be answered in 2 ways (yes or no)Therefore, no. of ways of answering 12 questions = 212 = 4096ways.Question:What is the value of 1.1! + 2.2! + 3!.3! + ............ n.n!,where n! means n factorial or n(n-1)(n-2)...1Solution:1.1! = (2 -1).1! = 2.1! – 1.1! = 2! - 1!2.2! = (3 - 1).2! = 3.2! - 2! = 3! - 2!3.3! = (4 - 1).3! = 4.3! - 3! = 4! - 3!......n.n! = (n+1 - 1).n! = (n+1)(n!) - n! = (n+1)! - n!Summing up all these terms, we get (n+1)! - 1!Question:In how many ways can the letters of the word MANAGEMENTbe rearranged so that the two As do not appear together?Solution:The word MANAGEMENT is a 10 letter word.Normally, any 10 letter word can be rearranged in 10! ways.However, as there are certain letters of the word repeating,we need to account for those. In this case, the letters A, M, Eand N repeat twice each.
Therefore, the number of ways in which the letters of theword MANAGEMENT can be rearranged reduces to .The problem requires us to find out the number of outcomesin which the two As do not appear together.The number of outcomes in which the two As appear togethercan be found out by considering the two As as one singleletter. Therefore, there will now be only 9 letters of whichthree of them E, N and M repeat twice. So these 9 letters with3 of them repeating twice can be rearranged in ways.Therefore, the required answer in which the two As do notappear next to each other =Total number of outcomes - the number of outcomes in whichthe 2 As appear together=> ways.**Question:There are 5 Rock songs, 6 Carnatic songs and 3 Indi pop songs.How many different albums can be formed using the aboverepertoire if the albums should contain at least 1 Rock songand 1 Carnatic song? (1)15624 (2)16384 (3)6144 (4) 240
Correct Answer - (1)Solution:There are 2n ways of choosing ‘n’ objects. For e.g. if n = 3,then the three objects can be chosen in the following 23 ways- 3C0 ways of choosing none of the three, 3C1 ways of choosingone out of the three, 3C2 ways of choosing two out of thethree and 3C3 ways of choosing all three.In the given problem, there are 5 Rock songs. We can choosethem in 25 ways. However, as the problem states that thecase where you do not choose a Rock song does not exist (atleast one rock song has to be selected), it can be done in 25 -1 = 32 - 1 = 31 ways.Similarly, the 6 Carnatic songs, choosing at least one, can beselected in 26 - 1 = 64 - 1 = 63 ways.And the 3 Indi pop can be selected in 23 = 8 ways. Here theoption of not selecting even one Indi Pop is allowed.Therefore, the total number of combinations = 31 . 63 . 8 =15624Question:How many words can be formed by re-arranging the letters ofthe word ASCENT such that A and T occupy the first and lastposition respectively? (1)5! (2)4!
(3)2! (4) 6! / 2!Correct Answer - (2)Solution:As A and T should occupy the first and last position, the firstand last position can be filled in only one way. The remaining4 positions can be filled in 4! Ways by the remaining words(S,C,E,N,T). hence by rearranging the letters of the wordASCENT we can form 1x4! = 4! words.**Question:A team of 8 students goes on an excursion, in two cars, ofwhich one can seat 5 and the other only 4. In how many wayscan they travel?Solution:There are 8 students and the maximum capacity of the carstogether is 9.We may divide the 8 students as followsCase I: 5 students in the first car and 3 in the secondOr Case II: 4 students in the first car and 4 in the secondHence, in Case I: 8 students are divided into groups of 5 and 3in 8C3 ways.Similarly, in Case II: 8 students are divided into two groups of4 and 4 in 8C4 ways.
Therefore, the total number of ways in which 8 students cantravel is 8C3 + 8C4 = 56 + 70 = 126.**Question:When four fair dice are rolled simultaneously, in how manyoutcomes will at least one of the dice show 3?Solution:When 4 dice are rolled simultaneously, there will be a total of64 = 1296 outcomes.The number of outcomes in which none of the 4 dice show 3will be 54 = 625 outcomes.Therefore, the number of outcomes in which at least one diewill show 3 = 1296 – 625 = 671Question:How many ways can 10 letters be posted in 5 post boxes, ifeach of the post boxes can take more than 10 letters?Solution:Each of the 10 letters can be posted in any of the 5 boxes.So, the first letter has 5 options, so does the second letterand so on and so forth for all of the 10 letters.i.e. 5.5.5….5 (upto 10 times)= 510.Here is another way to calculate it:
Including "none" as an option, there are 6 choices of appetizer, 11choices of main meal and 5 choices of dessert. Thus the total numberof choices is 6x11x5=330.One of these is not a meal though (no appetizer, no main meal and nodessert), so there are 329 possible meals.Circular permutationsThere are two cases of circular-permutations:-(a) If clockwise and anti clock-wise orders are different, then total number of circular-permutations is given by (n- 1)!(b) If clock-wise and anti-clock-wise orders are taken as not different, then total number of circular-permutations is given by (n-1)!/2!Proof(a):(a) Let‟s consider that 4 persons A,B,C, and D are sitting around a round tableShifting A, B, C, D, one position in anticlock-wise direction, weget the following agreements:-
Thus, we use that if 4 persons are sitting at a round table, thenthey can be shifted four times, but these four arrangements willbe the same, because the sequence of A, B, C, D, is same. Butif A, B, C, D, are sitting in a row, and they are shifted, then thefour linear-arrangement will be different.Hence if we have „4‟ things, then for each circular-arrangementnumber of linear-arrangements =4Similarly, if we have „n‟ things, then for each circular –agreement, number of linear – arrangement = n.Let the total circular arrangement = pTotal number of linear–arrangements = n.pTotal number of linear–arrangements= n. (number of circular-arrangements)Or Number of circular-arrangements = 1 (number of lineararrangements)n = 1( n!)/ncircular permutation = (n-1)!Proof (b) When clock-wise and anti-clock wise arrangements are not different, then observation can be made from both sides, and this will be the same. Here two permutations will be counted as one. So total permutations will be half, hence in this case. Circular–permutations = (n-1)!/2
Note: Number of circular-permutations of „n‟ different thingstaken „r‟ at a time:- (a) If clock-wise and anti-clockwise orders are taken as different, then total number of circular-permutations = n Pr /r (b) If clock-wise and anti-clockwise orders are taken as not different, then total number of circular – permutation = n Pr/2rExample: How many necklace of 12 beads each can be made from 18 beads of different colours?Ans. Here clock-wise and anti-clockwise arrangement s aresame. 18Hence total number of circular–permutations: P12/2x12= 18!/(6 x 24) Restricted – Permutations(a) Number of permutations of „n‟ things, taken „r‟ at a time, when a particular thing is to be always included in each arrangement= r n-1 Pr-1(b) Number of permutations of „n‟ things, taken „r‟ at a time, when a particular thing is fixed: = n-1 Pr-1(c) Number of permutations of „n‟ things, taken „r‟ at a time, when a particular thing is never taken: = n-1 Pr.
(d) Number of permutations of „n‟ things, taken „r‟ at a time, when „m‟ specified things always come together = m! x ( n- m+1) !(e) Number of permutations of „n‟ things, taken all at a time, when „m‟ specified things always come together = n ! - [ m! x (n-m+1)! ]Example: How many words can be formed with the letters ofthe word „OMEGA‟ when: (i) „O‟ and „A‟ occupying end places. (ii) „E‟ being always in the middle (iii) Vowels occupying odd-places (iv) Vowels being never together.Ans.(i) When „O‟ and „A‟ occupying end-places => M.E.G. (OA) Here (OA) are fixed, hence M, E, G can be arranged in 3!waysBut (O,A) can be arranged themselves is 2! ways. => Total number of words = 3! x 2! = 12 ways.
ii) When „E‟ is fixed in the middle => O.M.(E), G.A. Hence four-letter O.M.G.A. can be arranged in 4! i.e 24 ways. (iii) Three vowels (O,E,A,) can be arranged in the odd- places (1st, 3rd and 5th) = 3! ways. And two consonants (M,G,) can be arranged in the even- place (2nd, 4th) = 2 ! ways => Total number of ways= 3! x 2! = 12 ways. (iv) Total number of words = 5! = 120!If all the vowels come together, then we have: (O.E.A.), M,GThese can be arranged in 3! ways.But (O,E.A.) can be arranged themselves in 3! ways.=> Number of ways, when vowels come-together = 3! x3!= 36 ways=> Number of ways, when vowels being never-together= 120-36 = 84 ways.
Number of Combination of „n‟ different things, taken „r‟ at a timeis given by:-n Cr= n! / r ! x (n-r)!Proof: Each combination consists of „r‟ different things, whichcan be arranged among themselves in r! ways.=> For one combination of „r‟ different things, number ofarrangements = r! For nCr combination number of arrangements: r n Cr n=> Total number of permutations = r! Cr ---------------(1)But number of permutation of „n‟ different things, taken „r‟ at atime= nPr -------(2)From (1) and (2) : n Pr = r! . nCr or n!/(n-r)! = r! . nCr n or Cr = n!/r!x(n-r)!Note: nCr = nCn-r n or Cr = n!/r!x(n-r)! and nCn-r = n!/(n-r)!x(n-(n-r))!
= n!/(n-r)!xr!Restricted – Combinations (a) Number of combinations of „n‟ different things taken „r‟ at a time, when „p‟ particular things are always included = n-p Cr-p. (b) Number of combination of „n‟ different things, taken „r‟ at a time, when „p‟ particular things are always to be excluded = n-pCrExample: In how many ways can a cricket-eleven be chosenout of 15 players? if (i) A particular player is always chosen, (ii) A particular is never chosen.Ans:(i) A particular player is always chosen, it means that 10 players are selected out of the remaining 14 players. 14=. Required number of ways = C10 = 14C4= 14!/4!x19! = 1365(ii) A particular players is never chosen, it means that 11 players are selected out of 14 players.
14=> Required number of ways = C11= 14!/11!x3! = 364(iii) Number of ways of selecting zero or more things from „n‟different things is given by:- 2n-1Proof: Number of ways of selecting one thing, out of n-things = nC1Number of selecting two things, out of n-things =nC2Number of ways of selecting three things, out of n-things =nC3 Number of ways of selecting „n‟ things out of „n‟ things = nCn=>Total number of ways of selecting one or more things out ofn different things= nC1 + nC2 + nC3 + ------------- + nCn = (nC0 + nC1 + -----------------nCn) - nC0 = 2n – 1 [ nC0=1] 1. Examples: 1. Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA. Note: AB and BA represent the same selection. 2. All the combinations formed by a, b, c taking ab, bc, ca. 3. The only combination that can be formed of three letters a, b, c taken all at a time is abc.
4. Various groups of 2 out of four persons A, B, C, D are: AB, AC, AD, BC, BD, CD. 5. Note that ab ba are two different permutations but they represent the same combination. 2. Number of Combinations: The number of all combinations of n things, taken r at a time is: n n! n(n - 1)(n - 2) ... to r factors Cr = = . (r!)(n - r!) r! Note: n i. Cn = 1 and nC0 = 1. n ii. Cr = nC(n - r) Examples: 11 (11 x 10 x 9 x 8) i. C4 = = 330. (4 x 3 x 2 x 1) 16 x 15 x 14 16 x 15 x 14 ii. 16C13 = 16C(16 - 13) = 16C3 = = = 560. 3! 3x2x11. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done? A.564 B. 645 C. 735 D.756 2. In how many different ways can the letters of the word LEADING be arranged in such a way that the vowels always come together? A.360 B. 480 5040 C. 720 D.
3 how many different ways can the letters of the word In.CORPORATION be arranged so that the vowels always come together? A B 810 1440 . . C D 2880 50400 . . 4 Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed? A B 210 1050 . . C D 25200 21400 . . 5. In how many ways can the letters of the word LEADER be arranged? A B 72 144 . . C D 360 720 . . 6 In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there? A B 159 194 . . C D 205 209 . . E None of these . 7 How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated? A B 5 10 . . C D 15 20 . . 8 In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women? A 266 B 5040
. .C D 11760 86400. .E None of these.9 A box contains 2 white balls, 3 black balls and 4 red balls. In howmany ways can 3 balls be drawn from the box, if at least one blackball is to be included in the draw?A B 32 48. .C D 64 96. .E None of these.10 In how many different ways can the letters of the word DETAILbe arranged in such a way that the vowels occupy only the oddpositions?A B 32 48. .C D 36 60. .E 120.11 In how many ways can a group of 5 men and 2 women be madeout of a total of 7 men and 3 women?A B 63 90. .C D 126 45. .E 135.12 How many 4-letter words with or without meaning, can be formedout of the letters of the word, LOGARITHMS, if repetition of lettersis not allowed?A B 40 400. .C 5040 D 2520
. .13 In how many different ways can the letters of the wordMATHEMATICS be arranged so that the vowels always cometogether?A B 10080 4989600. .C D 120960 None of these. .14 In how many different ways can the letters of the word OPTICALbe arranged so that the vowels always come together?A B 120 720. .C D 4320 2160. .E None of these.Answer:1 Option DExplanation:We may have (3 men and 2 women) or (4 men and 1 woman) or (5men only). Required number of ways= (7C3 x 6C2) + (7C4 x 6C1) + (7C5) 7x6x5 6x5 = x + (7C3 x 6C1) + (7C2) 3x2x1 2x1 7x6x5 7x6 = 525 + x6 + 3x2x1 2x1 = (525 + 210 + 21) = 756.Answer:2 Option CExplanation:
The word LEADING has 7 different letters.When the vowels EAI are always together, they can be supposed toform one letter.Then, we have to arrange the letters LNDG (EAI).Now, 5 letters can be arranged in 5! = 120 ways.The vowels (EAI) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720.Answer: 3 Option DExplanation:In the word CORPORATION, we treat the vowels OOAIO as oneletter.Thus, we have CRPRTN (OOAIO).This has 7 (6 + 1) letters of which R occurs 2 times and the rest aredifferent. 7!Number of ways arranging these letters = = 2520. 2!Now, 5 vowels in which O occurs 3 times and the rest are different,can be arranged 5!in = 20 ways. 3! Required number of ways = (2520 x 20) = 50400Answer:4 Option CExplanation:
Number of ways of selecting (3 consonants out of 7) and (2 vowelsout of 4) = (7C3 x 4C2) 7x6x5 4x3 = x 3x2x1 2x1 = 210.Number of groups, each having 3 consonants and 2 vowels = 210.Each group contains 5 letters.Number of ways of arranging = 5!5 letters among themselves =5x4x3x2x1 = 120. Required number of ways = (210 x 120) = 25200.Answer: 5Option CExplanation:The word LEADER contains 6 letters, namely 1L, 2E, 1A, 1D and1R. Required number of ways =6!=Answer:6 Option DExplanation:We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boysand 1 girl) or (4 boys). Required number 6 = ( C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)of ways = (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2) 6x5 4x3 6x5x4 6x5 = (6 x 4) x + x4 + 2x1 2x1 3x2x1 2x1 = (24 + 90 + 80 + 15)
= 209.Answer:7 Option DExplanation:Since each desired number is divisible by 5, so we must have 5 at theunit place. So, there is 1 way of doing it.The tens place can now be filled by any of the remaining 5 digits (2,3, 6, 7, 9). So, there are 5 ways of filling the tens place.The hundreds place can now be filled by any of the remaining 4digits. So, there are 4 ways of filling it. Required number of numbers = (1 x 5 x 4) = 20Answer:8 Option C Explanation:Required number of ways= (8C5 x 10C6) = (8C3 x 10C4) 8 x 7 x 6 10 x 9 x 8 x 7 = 3 x 2 x 1x 4 x 3 x 2 x 1 = 11760.Answer9: Option CExplanation:We may have(1 black and 2 non-black) or (2 black and 1 non-black)or (3 black). Required number of ways= (3C1 x 6C2) + (3C2 x 6C1) + (3C3) 6x5 3x2 = 3x + x6 +1 2x1 2x1 = (45 + 18 + 1) = 64.
Answer:10 Option CExplanation:There are 6 letters in the given word, out of which there are 3 vowelsand 3 consonants.Let us mark these positions as under: (1) (2) (3) (4) (5) (6)Now, 3 vowels can be placed at any of the three places out 4, marked1, 3, 5.Number of ways of arranging the vowels = 3P3 = 3! = 6.Also, the 3 consonants can be arranged at the remaining 3 positions.Number of ways of these arrangements = 3P3 3! = 6.Answer:11 Option AExplanation: 7x 7 3 7 3Required number of ways = ( C5 x C2) = ( C2 x C1) 6 x = = 2x 3 63. 1Answer:12 Option CExplanation:LOGARITHMS contains 10 different letters.Required number of = Number of arrangements of 10 letters, takingwords 4 at a time. = 10P4 = (10 x 9 x 8 x 7) = 5040.
Answer:13 Option CExplanation:In the word MATHEMATICS, we treat the vowels AEAI as oneletter.Thus, we have MTHMTCS (AEAI).Now, we have to arrange 8 letters, out of which M occurs twice, Toccurs twice and the rest are different. 8! Number of ways of arranging these letters = = 10080. (2!)(2!)Now, AEAI has 4 letters in which A occurs 2 times and the rest aredifferent. 4!Number of ways of arranging these letters = = 12. 2! Required number of words = (10080 x 12) = 120960Answer:14 Option BExplanation:The word OPTICAL contains 7 different letters.When the vowels OIA are always together, they can be supposed toform one letter.Then, we have to arrange the letters PTCL (OIA).Now, 5 letters can be arranged in 5! = 120 ways.The vowels (OIA) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720.
ASSIGNMENT Question 1 The principal wants to arrange 5 students on theplatform such that the boy SALIM occupies the second positionand such that the girl SITA is always adjacent to the girl RITA .How many such arrangements are possible?Question 2 When a group- photograph is taken, all the seventeachers should be in the first row and all the twenty studentsshould be in the second row. If the two corners of the second roware reserved for the two tallest students, interchangeable only b/wthem, and if the middle seat of the front row is reserved for theprincipal, how many such arrangements are possible?Question 3 If there are six periods in each working day of aschool, in how many ways can one arrange 5 subjects such thateach subject is allotted at least one period?Question 4 How many numbers greater than 4,00,000 can beformed by using the digits 0, 2, 2, 4, 4, 5?**Question 5 The letters of the word ZENITH are written in allpossible order. How many words are possible if all these wordsare written out as in a dictionary? What is the rank of the wordZENITH?**Question 6 How many natural numbers not exceeding 4321 canbe formed with the digits 1,2,3 and 4 if the digits can repeat?**Question 7 Three married couples are to be seated in a rowhaving six seats in a cinema hall. If spouses are to be seated nextto each other, in how many ways can they be seated? Find also thenumber of ways of their seating if all the ladies sit together.Question8 A boy has three library tickets and 8 books of hisinterest in the library. Of these 8, he does not want to borrowChem. part II, unless Chem. Par I is also borrowed. In how manyways can be choose the three books to be borrowed?Question 9 A polygon has 44 diagonals. Find the number of itssides.** Question10 In how many ways can 10 things be equally dividedb/w (i) two persons (ii) two heaps ?**Question11 In an exam. there are three multiple choicequestions and each question has 4 choices. Find the no. Of ways inwhich a student fails to get all answer correct.
Question12 In how many ways can the letters of the wordPERMUTATIONS be arranged if the(i) Words start with P and end with S? (ii) Vowels are alltogether ? (iii) T’s are together? (iv) There is no restriction? (v) There are always 4 letters b/w P & S?Question 13 Find the no. Of perm. Of n things taken r at a time inwhich 2 given things always occur. In how many pers. , they arealways excluded?**Question 14 There are 10 points in a plane , no three of whichare in the same st. Line excepting 4 points, which are collinear.Find the (i) No. Of st. Lines obtained from the pairs of these points , (ii)no. Of ∆ that can be formed with the vertices as these points.Question 15 Find the no. of per. Of n different things taken r at atime such that two specific things occur together.ANSWERS ( WITH HINTS)Answer 1 since the boy SALIM occupies the second position, wehave to arrange the remaining 4 students, according to conditiontwo seats III, IV OR IV , V May be occupied by SITA & RITA in 4 ways, thenremaining seat may be occupied by 5th student in 1 way only.Sono. Of arrangements = 2.4.1=8.Answer 2 The remaining 6 teachers can be arranged in the frontrow in 6! Ways. (∵ middle seat is reserved for principal). Theremaining 18 students can be arranged in the second row in 18!Ways. (two corner are reserved for 2 tallest , they can be occupied2 ways) ∴ total no. Of ways = 6!.(18)!.2!Answer 3 Five subjects can be allotted 5 periods out of the sixperiods in 6P5 ways. Now one period is left and it can be allotted toany one of the 5 subjects in 5 ways. So, total no. Of ways = 6P5 . 5= 3600.Answer 4 When 5 occurs the extreme left position(greater than 4) ,then no. Of ways = = 30. When 4 occurs at extreme left , then
no. Of ways = 5!/2!=60 ( 0,2,2,4,5[2 appears twice]) ∴ total no. Ofways 90.Answer 5 (i) total no. Of possible words = 6!=720 (ii) The no. Of words beginning with E = 5! = 120 , similarly forH ,I, N or T is 120. The words start with Z will also be 120 and will have theirrank from 601 to 720.Of these 120 words start with Z, the no. Ofwords with E in the second place is 120/5 = 24. ZE**** will havetheir rank from 601 to 624. Of these 24 words , NO. Of wordswith H(or I or N or T) in the third place is 24/2=6 Rank order : ZEH*** , ZEI***, ZEN*** 601 to 606, 607 to612, 613 to 618 resp., start with ZEN are 613...ZENHIT, 614...ZENHTI, 615... ZENIHT, 616... ZENITH.Answer 6 1- digit nos. = 4, 2-digit nos. = 4x16, 3-digit nos. = 64,4-digit nos. 256. Out of 256 , let us discard those which are greaterthan 4321. (i) 4 at thousand’s place & hundred’s place= 22.214.171.124=16 , 4 4 1 1 2 2 3 3 4 4Similarly (ii) 4 at thousand’s place ,3 at hundred’s place, 3 or 4 inten’s = 126.96.36.199=8, (iii) 4 at thousand’s place ,3 at hundred’s place, 2 in ten’s & 2,3,4at unit’s= 188.8.131.52=3. Total 4-digit no.greater than 4321 are = 27,answer is 256-27=313.Answer 7(i) Let A, B, C be three married couples can be seated =3! & can sit in 2 ways =2! ∴ req. Ways of seating = 3!.2!.2!.2!=48 . (ii) three ladies can be seated in 4 ways which are at seatnumber (1,2,3), (2,3,4), (3,4,5), (4,5,6). They can interchange theirseats =3! Ways. Men can be seated at three remaining seats in 3! ,so req. No. Of ways = 4.3!.3!= 144 .Answer 8 case-1 he borrows chem.. part II, chem.. part I and onemore book out of the remaining 6 (8-2) books.
Case-2 he does not borrow chem.. part-II and , soborrow all 3 books out of 7 ∴ total no. Of ways = 6C1 + 7C3=41.Answer 9 No. Of diagonals = nC2 - n = 44 ( nC2 = no.of st. Linesof polygon n sides).Answer 10 (i) 10 things be equally divided b/w two persons,groups are distinct , 10!/ (5!)2 = 252 (ii) 10 things be equallydivided b/w two heaps no distinction can be made = ½ (252)=126. Answer 11 Req. No. Of ways= 4.4.4 – 1=63.Answer 13 (i) 10!/2! (ii) consider 5 vowels as one letter (8!/2!).(5!) (iii) consider 2 T’s as one letter = (11)! (iv) 12!/2! (v) 12 letters in 12 places as P****S****** leaving 4 places inbetween , Thus P & S may be filled up in 7 ways, same for S & P ,remaining 10 letters are arranged in 10!/2! Ways ∴ req. No. ofways 10!/(3!.2!.2!) = 151200. Answer 14 No. Of per. Of n things taken r at a time is same as theno. Of ways of filling up r places with n things, we arrange 2things which can be arranged in r places in rP2 ways. Alsoremaining (r-2) places can be filled up with remaining (n-2) thingsin n-2Pr-2 ways.No. Of per. (2 are included) = rp2 . n-2Pr-2 , No. Of per. (2 are notincluded) = n-2PrAnswer 15 (i) No. Of st. Lines formed by joining 10 points ,taking 2 at a time = 10C2 = 45. No. Of st. Lines joining 4 points =4 C2 = 6 , but 4 are collinear , when join pairwise give only one line.Req. No. of lines = 45 – 6 + 1=40 (ii) No. of ∆s formed by joining the points , taken 3 at a time =10 C3 = 120, no. of ∆s formed by joining 4 points , taken 3 at a time= 4C3 = 4 Req. No. of ∆s = 120 – 4 = 116.Answer 16 Two specific things can be in r places in (r-1) ways andtwo things can be arranged among themselves in 2! Ways and theremaining (n-2) things will be arranged in (r-2) places in n-2Pr-2ways. Req. Per. = 2! (r-1) . n-2Pr-2 .