Permutation and Combination

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Permutation and Combination Permutation implies arrangement where order of things is important. It includes various patterns like word formation, number formation, circular permutation etc. Combination means selection where order is not important. It involves selection of team, forming geometrical figures, distribution of things etc. Factorial = Factorial are defined for natural numbers, not for negative numbers. n! = n.(n-1).(n-2)... 3.2.1 For example: 1) 4! = 4.3.2.1 = 24 2) 6! 4! = 6.5.4! 4! = 6.5 1 = 30 3) 0! = 1 PERMUTATION Implies Arrangement Order of things is important Permutation of three things a, b and c taking two at a time are ab, ba, ac, ca,bc and cb (Order is important). COMBINATION Implies Selection Order of things is NOT important Combination of three things a,b and c taking two at a time are ab, ca and cb (Order is not important). npr =! ( )! ncr =! ( )! r! npn = n! ncn = 1 N p0 = 1 N C0= 1 Example of Word Formation: BANKERSWAY.COM

Example - 1: How many new words can be formed with the word "PATNA"? Solution: In word "PATNA", P,T,N occurs once and A occurs twice. ****Always remember in word formation, if word repeats, number of repetition will be in denominator. So, total number of words that can be formed = 5! 2! = 60 i.e. Therefore, except PATNA there are 59 new words (60-1). Example 2 : How many words can be formed from the letters of the word "EXAMINATION"? Solution: E, X, M, T, O : Occurs ONCE A, I, N : Twice So, total number of words = 2 BANKERSWAY.COM 11! 2! 2! 2! (Total number of letters=11 and 3 letters are occurring twice) Problems for practice Problem 1: Choose permutation or combination in following terms:- 1) Selection of captain and bowler for a play. Permutation 2) Selection of four students for a lecture. Combination 3) Assigning people to their seats during conference. Permutation Problem 2: Evaluate 7P2. 4P3 Solution: [ 7! ] [ 4! ] 5! 1! (7 6). (4 3 2) 1008 Problem 3: Evaluate 5C2. 3C2 Solution: [ 5! ] [ 3! ] 3! 2! [ 5 4 ]. [3] 2! 1! 30 Problem 4: How many ways are there in selecting 5 members from 6 males and 5 females, consisting 3 males and 2 females? Solution: This is a case of combination i.e. selecting 3 males from 6 males and 2 females from 5 females. Required number of ways = ( 6 5 ) 6! 5! 3! 3! 2! 3! C3 C2

[ 6 5 4 ] [ 5 4 ] 200 3 2 2 Problem 5: How many words can be formed by using letters of the word "DAUGHTER" so that the vowels come together? Solution: This is a case of permutation. In a word "DAUGHTER", there are 8 letters including 3 vowels (AUE) According to the question, vowels should always come together. Therefore, in this case we will treat all the vowels as one entity or one alphabet. This implies, in total there are 6 words (one word which is a group of vowels) These 6 words can be arranged in 6 ways P6 = = 6! = 720 WAYS 1! Also, three vowels in a group may be arranged in 3! ways 3! = 6 ways 6 6! Therefore, required number of words = (720 6) = 4320 P6 BANKERSWAY.COM PROBLEMS WITH SOULATION 1) 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 e) None of these 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 c) 720 d) 5040 e) None of these 3) In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together? a) 810 b) 1440 c) 2880 d) 50400 e) None of these 4) Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed? a) 210 b) 1050 c) 25200 d) 21400 e) None of these 5) In how many ways can the letters of the word 'LEADER' be arranged? a) 72 b) 144 c) 360 d) 72 e) None of these

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) 159 b) 194 c) 205 d) 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) 5 b) 10 c) 15 d) 20 e) None of these 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) 1176 d) 86400 e) None of these 9) A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw? a) 32 b) 48 c) 64 d) 96 e) None of these 10) In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions? a) 32 b) 48 c) 36 d) 60 e) None of these 11) In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women? a) 63 b) 90 c) 126 d) 145 e) None of these 12) In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together? a) 10080 b) 4989600 c) 120960 13) In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together? a) 120 b) 720 c) 4320 d) 2160 e) None of these 14) A letter lock consists of 4 rings, each ring contains 9 non-zero digits. This lock can be opened by setting a 4 digit code with the proper combination of each of the 4 rings Maximum how many codes can be formed to open the lock? a) 4 9 b) 9p4 c) 9 4

15) How many different words can be made using the letters of the word HALLUCINATION if all consonants are together? a) 129780 b) 1587600 c) 35600 16) If all S s come together, then in how many ways the letters of the word SUCCESSFUL be arranged? a) 10080 b) 40080 c) 2378 17) In how many different ways can 6 different balls be distributed to 4 different boxes, when each box can hold any number of ball? a) 2048 b) 1296 c) (24) 2 d) 4096 e) None of these 18) What is the total number of 4 digit numbers that can be formed using the digits 0, 1, 2, 3, 4, 5 without repetition, such that the number is divisible by 9? a) 36 b) 28 c) 15 d) 18 e) None of these 19) Seven delegates are to address a meeting. If a particular speaker is to speak before another particular speaker, find the number of ways in which this can be arranged. a) 1220 b) 2520 c) 3250 d) 7826 e) None of these 20) In how many ways can 15 billiard balls be arranged in a row if 3 are red, 4 are white and 8 are black? a) 12 b) 18 c) 96 21) In how many ways can 4 books be arranged out of 16 books on different subjects? a) 34650 b) 43680 c) 43890 22) Four dice are rolled. The number of possible outcomes in which atleast o ne die shows 4 is : a) 671 b) 168 c) 176 23) In how many ways can the letters of the word APPLE be arranged? a) 720 b) 120 c) 60 d) 180 e) None of these 24) In how many different ways can the letters of the word RUMOUR be arranged? a) 180 b) 90 c) 30 d) 720 e) None of these

25) A number plate of a vehicle has always a fixed code UP-32 for Lucknow city followed by the number of particular vehicle which is in two parts. First part is occupied by 2 English alphabets and second part is occupied by 4 digit numbers (0001, 0002,. 9999). If the latest registration number of vehicle [UP-32-SK- 0123] find the number of vehicles registered before this vehicle number in Lucknow. a) 2449744 b) 4779644 c) 4669235 d) 9235888222 e) None of these 26) The number of positive integral solutions of abc = 42 is : a) 17 b) 27 c) 21 d) 3! 42 e) None of these 27) A four digit number is formed with the digits 1, 3, 4, 5 without repetition. Find the chance that the number is divisible by 5 : a) 3 4 d) 1 16 b) 1 4 e) None of these c) 9 16 28) 20 girls, among whom are A and B sit down at a round table. The probability that there are 4 girls between A and B is : a) 17 19 d) 6 19 b) 2 19 e) None of these c) 13 19 29) The probability that the birthdays of 4 different persons will fall in exactly two calendar months is : a) 77 b) 17 c) 11 1728 87 144 30) A committee of five persons is to be chosen from a group of 9 people. The probability that a certain married couple will either serve together or not at all is : a) 4/9 b) 5/9 c) 13/18 Data inadequate d) e) None of these 31) In how many different ways can the letters of the word SOFTWARE be arranged in such a way that the vowels always come together? a) 120 b) 360 c) 1440 d) 13440 e) 720 32) In how many different ways can the letters of the word AUCTION be arranged in such a way that the vowels always come together? a) 30 b) 48 c) 144 d) 576 e) None of these

33) In how many ways can 21 books on English and 19 books on Hindu be placed in a row on a shelf so that two books on Hindi may not be together? a) 3990 b) 1540 c) 1995 d) 3672 e) None of these 34) Two numbers a and b are chosen at random from the set of first 30 natural numbers. The probability that 2-2 is divisible by 3 is : a) 37 87 b) 47 87 c) 17 29 35) From a pack of 52 cards, two are drawn one by one without replacement. Find the probabilities that both of them are kings. a) 11 21 d) 1 121 b) 13 121 e) None of these c) 1 221 Solutions 1. Option D We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only) Required number of ways = ( 7 C 3 x 6 C 2) + ( 7 C 4 x 6 C 1) + ( 7 C 5) = 7 x 6 x 5 6 x 5 x 3 x 2 x 1 2 x 1 + ( 7 C 3 x 6 C 1) + ( 7 C 2) = 525 + = (525 + 210 + 21) = 756 7 x 6 x 5 x 6 + 7 x 6 3 x 2 x 1 2 x 1 2. Option C The word 'LEADING' has 7 different letters. When the vowels EAI are always together, they can be supposed to form one letter. Then, we have to arrange the letters LNDG (EAI). Now, 5 (4 + 1 = 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 3. Option D In the word 'CORPORATION', we treat the vowels OOAIO as one letter. Thus, we have CRPRTN (OOAIO).

This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different. Number of ways arranging these letters = 7! = 2520. 2! Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in 5! = 20 ways. 3! Required number of ways = (2520 x 20) = 50400 4. Option C Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = ( 7 C 3 x 4 C 2) = = 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 5. Option C = 5 x 4 x 3 x 2 x 1 = 120. Required number of ways = (210 x 120) = 25200 The word LEADER contains 6 letters, namely 1L, 2E, 1A, 1D and 1R. 6! Required number of ways = = 360 (1!)(2!)(1!)(1!)(1!) 6. Option D 7 x 6 x 5 x 4 x 3 3 x 2 x 1 2 x 1 We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys). Required number of ways = ( 6 C 1 x 4 C 3) + ( 6 C 2 x 4 C 2) + ( 6 C 3 x 4 C 1) + ( 6 C 4) = ( 6 C 1 x 4 C 1) + ( 6 C 2 x 4 C 2) + ( 6 C 3 x 4 C 1) + ( 6 C 2) = (6 4) + [ 6 5 4 3 ] + [ 6 5 4 4] +[ 6 5 ] 2 1 2 1 3 2 1 2 1 = (24 + 90 + 80 + 15)

= 209 7. Option D Since each desired number is divisible by 5, so we must have 5 at the unit 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 4 digits. So, there are 4 ways of filling it. Required number of numbers = (1 5 4) = 20 8. Option C Required number of ways = ( 8 C 5 x 10 C 6) = ( 8 C 3 x 10 C 4) = [ 8 7 6 3 2 1 10 9 8 7 4 3 2 1 ] = 11760 9. Option C We may have (1 black and 2 non-black) or (2 black and 1 non-black) or (3 black) Required number of ways = ( 3 C 1 x 6 C 2) + ( 3 C 2 x 6 C 1) + ( 3 C 3) [ ] [ = 3 6 5 2 1 + 3 2 2 1 6] + 1 = (45 + 18 + 1) = 64 10. Option C There are 6 letters in the given word, out of which there are 3 vowels and 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, marked 1, 3, 5 Number of ways of arranging the vowels = 3 P3 = 3! = 6 Also, the 3 consonants can be arranged at the remaining 3 positions.

Number of ways of these arrangements = 3 P 3 = 3! = 6 Total number of ways = (6 6) = 36 11. Option A Required number of ways = ( 7 C 5 x 3 C 2) = ( 7 C 2 x 3 C 1) = [ 7 6 3] = 63 12. Option C 2 1 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, T occurs twice and the rest are different. Number of ways of arranging these letters = (2!)(2!) = 10080 Now, AEAI has 4 letters in which A occurs 2 times and the rest are different. Number of ways of arranging these letters = 4! (2!) = 12 Required number of words = (10080 12) = 120960 13. Option B The word OPTICAL contains 7 different letters. When the vowels OIA are always together, they can be supposed to form 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 6) = 720 14. Option C 9 9 9 9 = 9 4 15. Option B H L C N T A U I O L N A I There are total 13 letters out of which 7 are consonants and 6 are vowels. Also there are 2L s, 2N s, 2A s and 2I s 7! If all the consonants are together then the number of arrangements = But the consonants can be arranged themselves in Hence the required number of ways = 8! 7! ways. (2!)(2!) 7! 7! (2!)(2!) (2!)(2!) (2!)(2!) 16. Option E = (1260) 2 = 1587600 S U C E F L

S U C S There are 10 letters in the word SUCCESSFUL and S occurs 3 times, U occurs 2 times and C occurs 2 times. The letters of the word SUCCESSFUL can be arranged in = 17. Option D 10! ways 3! 2! 2! = 151200 ways Every ball can be distributed in 4 ways. Hence the required number of ways = 4 4 4 4 4 4 = 4 6 = 4096 18. Option A There are two sets of numbers 0, 1, 3, 5 and 0, 2, 3, 4. Therefore number of 4 digit numbers using digits 0, 1, 3, 5 = 3 3 2 1 = 18 Similarly number of 4 digit numbers using digits 0, 2, 3, 4 = 3 3 2 1 = 18 Hence the total required numbers = 18 + 18 = 36 19. Option B Total number of ways = 7! Let A has to speak before B. Now since there are half of the total cases in which A speaks before B (similarly in half of the total cases B speaks before A) Required number of ways = 1 7! = 2520 2 20. Option C Required number of ways = 15! 3! 4! 8! 21. Option B = 96 16p4 = 43680 22. Option A Total number of possible outcomes = 6 4 The number of possible outcomes in which 4 does not appear on any die is 5 4. Therefore the number of possible outcomes in which atleast one die shows digit 4 = 6 4-5 4 = 671

23. Option C The word APPLE contains 5 letters, 1A, 2P, 1L and 1E. 5! Required number of ways = = 60 (1!) (2!) (1!) (1!) 24. Option A The word RUMOUR contains 6 letters, namely 2R, 2U, 1M and 1U. 6! Required number of ways = = 180 (2!) (2!) (1!) (1!) 25. Option B Number of vehicles registered upto RZ 9999 = 18 26 (9999) Number of vehicles registered between SA-0001 and SJ 9999 = 1 10 9999 Therefore number of vehicles registered before SK-0123 = 18 26 9999 + 1 10 9999 + 122 = 9999 478 + 122 = 4779644 26. Option B 42 = 2 3 7 Here each of a, b and c can take 3 values. Hence the required number of solutions = 3 3 3 = 27 27. Option B Total possible number of 4 digits = 4! = 24 The number is divisible by 5 if unit digit itself is 5. Therefore we fix 5 at unit place and then remaining 3 places can be filled up in 3! Ways. Hence, the required probability = 3! 4! = 6 = 1 24 4 28. Option B 20 girls can be seated around a round table in 19! Ways. So, exhaustive number of cases = 19! Excluding A and B, out of remaining 18 girls, 4 girls can be selected 18 C 4 ways which can be arranged in 4! Ways. Remaining 20 - (4-2) = 14 girls can be arranged in 14! Ways. Also A and B mutually can be arranged in 2! Ways. Required number of arrangements = 18 C 4 4! 2! 14! = 18! 2 Required probability = 18! 2 = 2 19! 19 29. Option A

Since a person s birthday can fall in any of the 12 months. So, total number of ways = 12 4 Now, any two months can be chosen in 12 C 2 ways. The 4 persons birthday can fall in these two months in 2 4 ways. Out of these 2 4 ways there are two ways when all of the four birthdays fall in one month. So, favourable number of ways = 12 C 2 (2 4-2) 30. Option A 12 4 = 77 1728 Total number of ways in which 5 people can be chosen out of 9 people = 9 C 5 = 126 Number of ways in which the couple serves the committee = 7 C 3 2 C 2 = 35 Number of ways in which the couple does not serve the committee = 7 C 5 = 21 Favourable number of cases = 35 + 21 = 56 Hence, the required probability = 56 = 4 126 9 31. Option E The word SOFTWARE contains 8 different letters. When the vowels OAE are always together, they can be supposed to form one letter. Thus, we have to arrange the letters SFTWR (OAE). Now, 5 letters can be arranged in 6! = 720 ways The vowels (OAE) can be arranged among themselves in 3! = 6 ways. Required number of ways = (720 6) = 4320 32. Option D The word AUCTION has 7 different letters. When the vowels AUIO are always together, they can be supposed to form one letter. Then,we have to arrange the letters CTN (AUIO). Now, 4 letters can be arranged in 4! = 24 ways. The vowels (AUIO) can be arranged among themselves in 4! = 24 ways. Required number of ways = (24 24) = 576 33. Option B In order that two books on Hindi are never together, we must place all these books as under: X E X E X E X. X E X Where E denotes the position of an English book and X that of a Hindi book. Since there are 21 books on English, the number of places marked X are therefore, 22 Now, 19 places out of 22 can be chosen in 22 C 19 = 22 C 3 = 22 21 20 = 1540 ways 3 2 1 Hence, the required number of ways = 1540

34. Option B Out of 30 numbers 2 numbers can be chosen in 30 C 2 ways. So, exhaustive number of cases = 30 C 2 = 435 Since 2-2 is divisible by 3 if either a and b are divisible by 3 or none of a and b is divisible by 3. Thus, the favourable numbers, of cases = 10 C 2 + 20 C 2 = 235 Hence, required probability = 235 = 47 35. Option C Required probability = 4 3 = 435 87 1 52 51 221