Permutations and Combinations Solved Problems

PERMUTATION AND COMBINATION Page # 3 SOLVED PROBLEMS Ex.1 If Sol. 1 x 1 + = , then find the value of x 10 ! 11! 9! 1 1 1 x x 1  = + =  9 ! 10! 10 . 9 ! 11 . 10 . 9 ! 11! 9! LM N 1 OP FG Q H x IJ K 1 x 1 1  9! 1  10 = 11 . 10 . 9 !  1  10 = 11.10 x 11  10 = 11.10  x = 11 . 11 = 121 Ex.2 Find the number of different words (meaningful or meaningless) can be formed by taking four different letters from English alphabets Sol. The first letter of four letter word can be chosen by 26 ways, second by 25 ways, third by 24 ways and fourth by 23 ways. So number of four letter words = 26 × 25 × 24 × 23 = 358800 Ex.3 If 56P r +6 : 54 P find the value of r 56 Sol. 54 Pr  6 Pr  3  r +3= 30800 : 1 then 30800 1 56 ! (30800) x 54 !  (56  r  6) ! = (54  r  3) ! Ex.6 Find how many numbers can be formed between 20000 and 30000 by using digits 2, 3, 5, 6, 9 when digits may be repeated? Sol. First digit between 20000 and 30000 will be 2 which can be chosen by one way. Every number will be of five digits and all the digits can be anything from the given five digit except first digit. So each digit of the remaining four digits can be chosen in 5 0ways  required numbers = 1 × 5 × 5 × 5 × 5 = 625 Ex.7 Find the number of three letters words can be formed from the letters of word ‘SACHIN’ when I do not come in any word Sol. There are 6 letters in the given word. Then the number of three letters words from the remaining 5 letters after removing I is = 5P3 = 5 × 4 × 3 = 60 Ex.8 Find the number of numbers lying between 100 and 1000 which can be formed with the digits 0, 1, 2, 3, 4, 5, 6 Sol. Required numbers will have 3 digits so their total number = 7P3 - 6P2 = 180 Ex.4 Find the number of ways in which 2 vacancies can be filled up by 13 candidates Sol. The no. of ways to fill up 2 vacancies by 13 candidates is13P = 13 × 12 = 156 2 Ex.9 Find how many numbers between 1000 and 4000 (including 4000) can be formed with the digits 0,1,2,3,4 if each digit can be repeated any number of times? Sol. Required number will have 4 digits and their thousand digit will be 1 or 2 or 3 or 4. The number of such numbers will be 124, 125, 125 and 1 respectively.  Total numbers = 124 + 125 + 125 + 1 = 375 Ex.5 Find how many different words beginning with A and ending with L can be formed by using the letters of the word’ ANILMANGAL’? Sol. After fixing the letters A and L in the first and last places, the total number of available places are 8 and the letters are also 8. Out of these 8 letters there are 2 groups of alike letters. Therefore no. of words 8! = =10080 2! 2! Ex.10 Find the number of ways in which 7 girls can be stand in a circle so that they do not have the same neighbour in any two arrangements? Sol. Seven girls can keep stand in a circle by (7  1) ! number of ways, because there is no 2! difference in anticlockwise and clockwise order of their standing in a circle. (7  1) !  = 360 2!  56 × 55 × (51–r) = 30800 30800  (51–r) = 56 x 55 = 10  r = 51 – 10 = 41 PERMUTATION AND COMBINATION Page # 4 Ex.14 Out of 10 white, 9 black and 7 red balls, find 5 47C Ex.11 Find Sol. + 4  52r C 3 the number of ways in which selection of r 1 The given expression can be written as one or more balls can be made. 5 Sol.  52r C 3 + 47C4 (10 + 1) (9 + 1) (7+ 1) – 1 r 1 = 51C 3 50C 3 + + [We know that = 51C = 51C = 51C = 51C 3 + 50C 3 + 50C 3 + 50C + 51C 3 49C + 48C + 47C + 47C 3 3 3 4 n+1C nC + nC ] r+1 r r+1 + 49C3 + 48C3 + 48C4 + 49C3 + 49C4 + 50C4 = 52C4 The required number of ways are 3 3 3 4 = 11 × 10 × 8 –1 = 879 Ex.15 Find the number of words which can be formed taking 4 different letters out of the letters of the word ‘ASSASSINATION’, Sol. Total No. of selections of 4 different letters = 6C4 Ex.12 A candidate is required to answer 6 out of  10 questions which are divided into two = groups each containing 5 questions and he Sol. 6C 4 . 4! is not permitted to attempt more than 4 Ex.16 There are four balls of different colours and from each group. Find the number of ways four boxes of colours same as those of the in which he can make up his choice balls. Find the number of ways in which the Let there be two groups A and B each balls, one each box, could be placed such containing 5 questions. Questions to be that a ball does not go to box of its own attempted is 6, but not more than 4 from colour any group. The candidate can select the Sol. Number of derangements are questions in following ways: (i) 4 from group A and 2 from group B. = 4! (ii) 3 from group A and 3 from group B. (iii) 2 from group A and 4 from group B 5C are × 4 5C 5 2, C3 × 5C 5 3, C 2 × 5C RS1  1  1  1  1 ...(1) 1 UV n! W T 1! 2! 3! 4! 4  Number of ways of selecting 6 questions 5C 4 × 5C 2+ 5C 3× 5C 3+ 5C 2 × 5C = 12 – 4 + 1 = 9 problems is given by n! n respectively. = RS 1  1  1 UV T 2! 3 ! 4 ! W {Since number of derangements in such a The number of selections in the above cases Ex.17 Find the number of 4 digit numbers divisible 4 by 5 which can be formed by using the = 50 + 100 + 50 = 200 Ex.13 Find in how many ways can a committee Sol. Total no. of different words digits 0, 2, 3, 4, 5 Sol. For a number to be divisible by 5, unit place consisting of one or more members be formed should be occupied out of 12 members of the Municipal (i) If unit place is 0 then remaining 3 places Corporation can be filled by Required number of ways = = 12C 1 12C 12 + = 12C 2 212 +12C 3 –1 4096 – 1 = 4095 +.....+ 4P 3 by 0 or 5ways = 24 (ii) If unit place is 5 then no. of ways = 4! – 3! = 18  Total number of ways = 24 + 18 = 42 PERMUTATION AND COMBINATION Page # 5 Ex.18 Find the sum of all 5 digit numbers which Sol. each containing two consonants and three can be formed using digits 1, 2, 3, 4, 5 vowels. Now, each group contains 5 letters Using formula (Here n = 5) which are to be arranged in such a way that Sum = (1+ 2 + 3 + 4 + 5) 4! (11111) 2 consonants occur together. Considering 2 = 15 × 24 × 11111 = 3999960 consonants as one letter, we have 4 letters which can be arranged in 4! ways. But two Ex.19 Find the number of diagonals in an octagon Sol. consonants can be put together in 2! ways. Here n = 8 (given) Therefore, 5 letters in each group can be The number of diagonals are given by n (n  3) 2 8 ( 8  3) 8. 5 = = 20  2 2 Ex.20 Out of 10 given points 6 are in a straight arranged in = line. Find the number of the triangles formed Sol. by joining any three of them A triangle can be formed by joining three points, so there will be 10C3 triangles joining any three out of 10 points. But 6 of these 10 points are collinear so these 6 points will give no triangle. Hence the required number of triangles =10C3 – 6C3 = 120 – 20 = 100 Ex.21 Find the number of ways in which 5 biscuits can be distributed among two children Sol. Each biscuit can be distributed in 2 ways. Therefore number of ways of distributing the biscuits = 25 = 32 Hence, the required number of words = ( 5C3 × 3C 2) × 4! × 2! = 1440 Ex.23 Find how many numbers can be formed with the digits 0,1,2, 3,4,5 which are greater than 3000? Sol. The numbers greater than 3000 may contain 4,5 or 36 digits. Now the 6 digit numbers can be formed by using all 6 given digits which will be 6P 6. These numbers include the numbers which starts with 0. Such type of numbers are 5P 5. Therefore 6 digit numbers greater than 3000 = 6P 6 – 5P 5 = 600 Similarly 5 digit numbers greater than 3000 = 6P 5 – 5P 4 = 600 Now 4 digit numbers which are greater than Now number of ways in which either of the 3000 should begin with the digit 3,4 or 5. If two children does not get any biscuit = 2. the first place of the number is occupied by  Required number of ways of distribution = 32 – 2 = 30 Ex.22 Find how many five-letter words containing 3 vowels and 2 consonants can be formed using the letters of the word ”EQUATION’ so that the two consonants occur together? Sol. 4! × 2! ways. There are 5 vowels and 3 consonants in the word ‘EQUATION’. Three vowels out of 5 and 2 consonants out of 3 can be chosen in 5C3 × 3C2 ways. So, there are 5C3 × 3C2 groups the digit 3, 4 or 5, then the remaining three places of the number can be filled in 60 ways. 5P 3 = Therefore numbers of 4 digits greater than 3000 = 3 x 60 = 180 Hence required numbers = 600 + 600 + 180 = 1380