Permutations and combinations
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Permutations and combinations

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This material is for PGPSE / CSE students of AFTERSCHOOOL. PGPSE / CSE are free online programme - open for all - free for all - to promote entrepreneurship and social entrepreneurship PGPSE is for ...

This material is for PGPSE / CSE students of AFTERSCHOOOL. PGPSE / CSE are free online programme - open for all - free for all - to promote entrepreneurship and social entrepreneurship PGPSE is for those who want to transform the world. It is different from MBA, BBA, CFA, CA,CS,ICWA and other traditional programmes. It is based on self certification and based on self learning and guidance by mentors. It is for those who want to be entrepreneurs and social changers. Let us work together. Our basic idea is that KNOWLEDGE IS FREE & AND SHARE IT WITH THE WORLD

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Permutations and combinations Permutations and combinations Presentation Transcript

  • PERMUTATIONS AND COMBINATIONS by : DR. T.K. JAIN AFTERSCHO ☺ OL centre for social entrepreneurship sivakamu veterinary hospital road bikaner 334001 rajasthan, india FOR – PGPSE / CSE PARTICIPANTS [email_address] mobile : 91+9414430763
  • My words..... My purpose here is to give a few questions on fundamentals of permutations and combinations. I welcome your suggestions. I also request you to help me in spreading social entrepreneurship across the globe – for which I need support of you people – not of any VIP. With your help, I can spread the ideas – for which we stand....
  • What is permutation & combination? Permutation = it denotes order / Sequence but combination = it only denotes that some objects are together example : ABC can have only one combination taking all of them together. But permutations are many : - ABC,ACB,BCA,BAC,CBA,CAB
  • What is formula of permutation ? Npr = n! / (n-r)! p=permutation n= total number of objects r=how many objects you are taking at a time ! = multiply with reducing numbers till it reaches 1 example : 5p5 = 5! / (5-5)! 5!=5*4*3*2*1 0! = 1 thus answer = 120 answer
  • There are 5 books on maths, 3 on entrepreneurship and 2 on economics. In how many ways can we arrange them. We have to keep books of a subject together. There are 3 subjects so we can arrange them in 3! methods = 6 there are 5 books on maths , we have 5! methods = 120 there are 3 books on entrepreneurship , so we have 3! methods there are 2 books on economics , so we have 2! methods. Total we have : 6*120*6*2 = 8640 answer
  • A court has given 6 to 3 decision in favaour of an issues. In how many ways can it reverse the decision? There are 9 judges, 5 will make a majority there may be 5,6,7,8,9 judges against the issue we have 9c5 or 9c6 or 9c7 or 9c8 or 9c9 methods : 126 + 84+36+9+1 =256 answer
  • Three persons go into a room which has 8 seats, in how many ways can they occupy the seats ? First person has 8 options, 2 nd person has 7 options and 3 rd person has 6 options. Total = 8*7*6 = 336 answer
  • There are 5 routes between Bikaner and Delhi. You may choose any route. In how many ways can you go and return from Delhi? 5*5 = 25
  • How many permutations are possible from ACCOUNTANT ? There are 10 digits, there are 2A, 2C, 2T, 2N, so we have : 10! / (2!*2!*2!*2!) answer
  • How many different 4 digit letters can you make out of A,B,C,D,E? N = 5 (A,B,C,D,E) R = 4 formula = Npr = n! / (n-r)! =5!/(5-4)! = 120 answer
  • How many different 4 digit numbers can you make out of 1,2,3,4,0? N = 5 (1,2,3,4,0) R = 4 but 0 cannot come in the first digit for first digit we have 4 options (1,2,3,4), for next digits, we can use 0. thus we have 4*4*3*2*1 = 96 options OR formula = Npr = n! / (n-r)! =5!/(5-4)! but this contains all those numbers which start with 0. so let us keep 0 as fixed for 1 st digit and solve it. Now we have to pick up 3 digit out of 4 contd.
  • contd..... If it is not 0, permutation will be : formula = Npr = n! / (n-r)! =5!/(5-4)! = 120 Zero fixed for 1 st potion, we have these options : Npr = n! / (n-r)! n=4,r=3 4!/(4-3)! = 24 deduct this 24 from 120 120 -24 = 96 answer you can use any formula (out of these 2), you get the same answer
  • How many different 4 digit numbers can you make out of 1,2,3,4,0 which are divisible by 2? Start with 96 of the last question now pick up all those which are ending with 1 : 3*3*2*1 = 18 similarly those which are ending with 3 3*3*2*1 = 18 thus 96 – (18+18) = 60 seems to be the answer
  • In how many ways can Raj invite any 3 of his 7 friends? This is a question of combination. Here order (sequence) is not important, his friends can come in any order. Thus this is a case of combination. Formula : N! / ((n-r)!*r!) you can calculate combination by dividing permutation by r! =7! / ((7-3)!*3!) =(7*6*5)/(3*2*1) = 35 answer
  • How many different words can you frame from FUTURE ? Here we have two U total we have 6 digits. Formula : N ! / L! N= total number of digits L = those digits which are repeated. Answer = 6! / 2! = 360 answer
  • How many different words can you frame from DALDA ? Here we have two D & A total we have 5 digits. Formula : N ! / L! N= total number of digits L = those digits which are repeated. Answer = 5! / (2!*2!) = 30 answer
  • In how many ways can 8 person sit around a round table ? For questions relating to round table , we have to use the following formula : (n-1)! So here answer = (8-1)! = 7! =5040 answer
  • How many 4 digit numbers can be formed out of 1,2,3,5,7,8,9 if no digit is repeated. Total number ofdigits = 7 formula = Npr n =7 r 4 7p4 = 7! / 3! =7*6*5*4 = 840
  • How many numbers greater than 2000 can be formed from 1,2,3,4,5. No repeatition is allowed. 5 digit numbers = 5! = 120 4 digit numbers,: we cant take 1 in the beginning. We have 4 options for 1 st digit 4 for 2 nd digit 3 for 3 rd digit ... 4*4*3*2*1 = 96 total = 216 answer
  • There are 6 books on english, 3 on maths, 2 on GK. In how many ways can they be placed in shelf, if books of 1 subject are together? We have 3 subjects so 3! books of same subjects can be interchanged. So answer : 3!*6!*3!*2! =6*720*6*2 = 51840 answer
  • How many words can we make out of DRAUGHT, the vowels are never separated? Number of vowels = 2 other digits = 5 we will treat vowels as 1 word so we have 6!. Vowels can be interchanged so 2! so answer = 6!*2! = 1440 answer
  • In how many ways can 8 pearls be used to form a necklace ? In questions of necklace, we use the following formula : ½ (N-1)! Here we can take reverse order of left to right or right to left, so divide by ½ =1/2 (8-1)! =2520
  • In how many number of ways can 7 boys form a ring ? (7-1) ! = 6! = 720 answer
  • 50 different jewels can be set to form necklace in how many ways ? ½ ( n -1) ! = ½ (50 -1)! =1/2 (49)!
  • How many number of different digits can be formed from 0,2,3,4,8,9 between 10 to 1000? Let us assume that repeatition is not allowed Let us make 2 digit numbers : for first digit we have 5 option, for 2 nd digit also we have 5 options (including 0) = 25 for 3 digit numbers : 5*5*4 = 100 total 125 if repeatition is allowed : for 2 digit : 5 * 6 = 30 for 3 digit : 5*6*6 = 180 total = 210 answer
  • What is the number of permutations of 10 different things taking 4 at a time in which one thing never comes ? = 9 p 4 = (9*8*7*6) =3024
  • There are 5 speakers (A,B,C,D,E) , in how many ways can we arrange their speach that A always speaks before B For A and then B without gap : Let us take A and B as one. 4! = 24 for A and then B let us keep B at 3 rd place and A at 1 st place =3! there are total 6 such possibilities so we have 6*6 = 36 total possibilities = 60 answer
  • 5 persons are sitting in a round table in such a way that the tallest person always sits next to the smallest person? Keep tallest and smallest person as 1. we have (4-1)! = 6 the tallest and the smallest person can be interchanged = 2 =12
  • How many words can be formed from MOBILE so that consonent always occupies odd place ? There are 3 odd and 3 even places. We have 3! *3! =36 answer
  • In how many ways can we arrange 6 + and 4 – signs so that no two – signs are together? + + + + + + there are 5 places between 2 +. one on extreme left and one on extreme right. We have 7 positions for – sign 7c4 we have 6 places for 6 + sign, so we have 6c6 total = 35 answer
  • There are 10 buses between Bikaner and Jaipur. In how many ways can Gajendra go to Jaipur and come back without using the same bus in return journey? There are 10 options while going there are 9 options while returning (one bus used earlier will not be used) 10*9 = 90 answer
  • In how many ways can yamini distribute 8 sweets to 8 persons provided the largest sweet is served to Jigyasha? 1 sweet is fixed so we have 7! = 5040 answer
  • Yamini & Jigyasha go to a train and they find 6 vacant seats. In how many ways can they sit? Yamini has 6 options but Jigyasha has only 5 options left = 6*5 = 30 answer
  • How many words can you make from DOGMATIC? 8! 40320 answer
  • Gajendra has 12 friends out of whom 8 are relatives. In how many ways can he invite 7 in such a way that 5 are relatives? 8c5 * 4c2 =56*6 =336 answer
  • There are 8 points on a plane. No 3 points are on a straight line. How many traiangles can be made out of these ? 8c3 = 56 answer
  • In how many ways can you form a committee of 3 persons out of 12 persons ? 12c3 =220 answer
  • How many different factors are possible from 75600 ? The factors are : 2^4* 3^3*5^2 *7 formula = (number of factors +1) (number of factors +1) .... - 1 (4+1)(3+1)(2+1)(1+1) -1 =119 answer
  • A box contains 7 red 5 white and 4 blue balls. How many selections can be made that we pick up 3 balls and all are red? It is a question of combination. Total possibilities = 7c3 7c3 = 7*6*5 / 3*2*1 = 35 thus there are 35 chances of getting
  • A box contains 7 red 5 white and 4 blue balls. What is the probability that in our selections we pick up 3 balls and all are red? Total possibilities for red = 7c3 7c3 = 7*6*5 / 3*2*1 = 35 total possibility of 3 balls : 16c3 =(16*15*14/3*2*1) =560 probability - thus there are 35/560 chances of getting red in all the three selections
  • What is the probability of getting 3 heads when I toss a coin 5 times? This is a case of binomial probability (where there are only 2 outcomes possible, we can use this theory) Here we can use this formula : Ncr (p)^r * (q)^(n-r) =n =5, p = ½ q = (1-p) = ½ , r = 3 5c3 (1/2)^3*(1/2)^2 =5/48 answer
  • In how many ways can Gajendra invite some or all of his 5 friends in party hosted by him? (at least 1) Frmula of combination of 1 to all = 2^n – 1 = 2^5 - 1 = 32-1 =31 answer
  • How many words can be formed by using all the letters of the word DRAUGHT so that a. vowels always come together & b. vowels are never together? A There are 2 vowels. We treat them as 1. solution : 6!*2! = 1440 answer b. total possibilities = 7! = 5040 number of cases when vowels are not together = 5040-1440 = 3600 answer
  • In how many ways can a cricket eleven be chosen out of a batch of 15 players. 15c11 =15! / ((15-11)!*11!) =15!/(4!*11!) =(15*14*13*12)/(4*3*2*1) 1365 answer
  • In how many a committee of 5 members can be selected from 6 men 5 ladies consisting of 3 men and 2 ladies 6c3 *5c2 =[(6*5*4)/(3*2*1)] [(5*4)/(2*1)] =20*10 =200 answer
  • How many 4-letter word with or without meaning can be formed out of the letters of the word 'LOGARITHMS' if repetition of letters is not allowed 10p4 =(10*9*8*7) =5040 answer
  • how many ways can the letter of word 'LEADER' be arranged We have two e, so divide 6p6 by 2 6!/2! =720 / 2 =360 answer
  • How many arrangements can be made out of the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together Let us treat all 4 vowels as 1 total digits are 11 we we take 11 – 4+1 = 8 digits vowels can be arranged among themselves = 4!/2! =8!/ (2!*2!) * 4!/2! = 120960 answer
  • In how many different ways can the letter of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions We have 3 odd and 3 even positions =3! *3! =36 answer
  • 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? Last digit must be 5 now we have 5 options for 1 st and 4 options for 2 nd digit =5*4 = 20 answer
  • In how many ways can 21 books on English and 19 books on Hindi be placed in a row on a self so that two books on Hindi may not be together? We have 22 places for Hindi books. 22p19 *21!
  • Out of 7 constants and 4 vowels how many words of 3 consonants and 2 vowels can be formed? Selection of 5 digits =7c3 *4c2 =35*6 = 210 5 digits can be arranged in 5! ways =120 total options : 210*120 = 25200 answer
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