Tips on cracking Aptitude Questions on Permutations

1. Tips on Solving Aptitude Type Questions on Permutation

2. Understanding Permutation & Combination Suppose we have 4 objects A, B, C and D and we are required to choose 3 from them and then arrange them on a shelf. This can be done in the following ways: Thus, there are 4 ways of choosing 3 objects from 4 and there are 6 ways of arranging the chosen objects. The process of selecting things is called combination and that of arranging things is called permutation. Examples of problems relating to combination: Formation of a team from a number of players. Formation of a particular committee from a number of players. Examples of problems relating to permutation: Arrangement of books on a shelf. Formation of numbers with the given digits. Formation of words with the given letters.

3. The Sum Rule The Sum Rule:If A and B are 2 disjoint events (A or B can occur but they never occur together) such that A occurs in m ways and B in n ways, then A or B can occur in m + n ways. Question: How many odd numbers less than 1000 can be formed using the digits 1,2,5,7,8 if repetition of digits is allowed? Solution: Total no. of digits= 5, No. of odd digits= 3. One-digit numbers: Only 3 1-digit numbers are possible: 1,5,7 Two-digit numbers: No. of possible ways= 5 x 3= 15 Three-digit numbers: No. of possible ways= 5 x 5 x 3= 75 Thus, there are (3+15+75) = 93 ways of forming odd numbers less 1000 using the given digits.

4. The Product Rule The Product Rule:If an operation can be performed in m ways, and if corresponding to each of these m ways of performing this operation, there are n ways of performing a second operation, then the number of ways of performing the two operations together is m x n. Question: There are 10 buses running between 2 towns X and Y. In how many ways can a man go from X to Y using a specific bus and return by a different bus? Solution: A man can go from X to Y in 10 ways and return in (10 – 1) ways= 9 ways. Thus, number of ways of doing the journey= 10 x 9 = 90.

5. Formula for Permutations Question: What is the number of ways in which 5 persons can occupy 3 vacant seats? Solution: Assume that any one of the 5 persons can sit in Seat 1. That would imply that the choice for filling Seat 1 can be done in 5 ways. We now have 4 persons left that can sit in Seat 2. That implies that the choice for filling Seat 2 can be done in 4 ways. With 3 persons left after Seat 1 & 2 are filled, we can fill Seat 3 in 3 ways. Applying product rule, no of ways of filling 3 seats with 5 persons = 5 x 4 x 3 = 60. Formula for Permutations: No of ways of choosing and arranging ‘r’ elements from a total of ‘n’ given elements is: nPr or P(n, r) = n! (n – r)! ? ? ? 5 persons 3 persons 4 persons

6. Restricted Permutations: Illustration 1 Question: In how many ways can a shelf for 4 books be formed out of 10 books such that a particular book is always (i) included (ii) left out? Solution: (i) If a book is always included, then it can come in the first, second, third or fourth positions (i.e.) it’s position can be selected in 4 ways. The other 3 books in the shelf can be selected from the remaining 9 books in P(9,3) ways Total Number of ways = P(9,3) x 4= 9! / 6! x 4 = 9 x 8 x 7 x 4 = 2016 (ii) If a book is always excluded, effectively we only have 9 books to fill in 4 positions. Total Number of ways = P(9,4)= 9! / 5!= 9 x 8 x 7 x 6= 3024

7. Restricted Permutations: Types and Formulae Restricted Permutations: Illustration 2 Question: In how many ways can 6 boys and 4 girls be arranged in a straight line such that no two girls are ever together? Solution: First, fixing the positions of boys, no. of permutations= 6! Since two girls cannot sit together, a girl can sit either between two boys or at the ends resulting in 7 possible seats as shown below: Fixing the positions of girls = P(7,4) Thus, no. of ways of arranging 6 boys and 4 girls such that no two girls are together = 6! x P(7,4)= 7x 6 x 5 x 4 x 6 x 5 x 4 x 3 x 2 x 1= 604800. G B1 G B2 G B3 G B4 G B5 G B6 G

8. Restricted Permutations: Illustration 3 Question: How many 5-digit numbers can be formed out of the digits 0, 1, 2,….., 9 if each number starts with 35 and no digit is repeated? Solution: Since the first two positions are defined and no digit is to be repeated, the remaining 3 positions have to be filled with digits 0,1,2,4,6,7,8,9, i.e., 8 digits. Number of ways of forming required number = P(8,3)= 8! / 5!= 8 x 7 x 6= 336. 3 5 ? ? ? 8 digits 6 digits 7 digits

9. Permutations of Alike Things The number of permutations (x) of n things taken all at a time where p things are alike of one kind, q things are alike of a second kind, r things are alike of a third king and so on, is, x= n! p! q! r! … Question: There are 3 copies each of 4 different books. Find the number of ways of arranging them on a shelf. Solution: Total no. of books= 3 x 4= 12 Required no. of ways of arranging them= 12! = 369600 3! 3! 3! 3! Question: Find the number of arrangements of the letters of the words ‘BANANA’ such that the 2 N’s do not appear together. Solution: There are 3 A’s and 2 N’s. Total no. of ways of arranging the letters= 6! / (3! 2!)= 60 No. of arrangements in which N’s appear together = 5! / 3! = 20. [We assume that the two N’s are combined to form a single character] Thus, no. of arrangements in which the 2 N’s do not appear together= 60 – 20= 40.

10. Permutations of Repeated Things The number of permutations of n different things taken r at a time, when each thing may occur any number of times is nr . Question: 8 different letters of the alphabet are given. Words of 4 letters are formed. Find the no. of such words with at least one letter repeated. Solution: If any letter can be used any no. of times, no. of letters that can be formed= 84 = 4096. No. of words with no letter repeated= P(8,4)= 8 x 7 x 6 x 5= 1680. No. of words with at least 1 letter repeated= 4096 – 1680= 2416. Question: How many 3-digit numbers can be formed with the digits 1,2,3,4,5 when the digits may be repeated? Solution: Required no. of 3-digit numbers that can be formed= 53 = 125. ? ? ? ? ? ? ? ? 8 letters 8 letters 8 letters 5 letters 8 letters 8 letters 7 letters 6 letters When letters cannot be repeated When letters can be repeated

11. Circular Permutations • Consider A, B and C to be arranged in a circular fashion. • 3 linear arrangements ABC, BCA and CAB are result in the same circular arrangement: • No of circular arrangements with n elements = No of linear arrangements with n elements/ n • = n! / n = (n – 1)! • The number of ways in which n objects taken r at a time can form a ring = nPr / r • If clock-wise and counter clock-wise arrangements are equivalent, divide the number of ways by 2 • Question: In how many ways can 10 boys and 5 girls sit around a circular table such that no 2 girls sit together? • Solution: 10 boys can be seated around a table in 9! ways. • There are 10 spaces between the boys which can be filled up by the 5 girls in P(10,5) ways. • Thus, total no. of ways of arranging the boys and girls= 9! x 10P5 = (9! 10! )÷ 5! • Question: Find the no. of ways in which 10 different flowers can form a garland so that 4 particular flowers are never separated. • Solution: Let the 4 flowers be considered as a single flower. Then we have 7 flowers. • These can be formed into a garland in 6! ways. • The 4 particular flowers can be arranged in 4! ways. • Thus, total no. of ways of forming the garland= (6! x 4! ) / 2 = 8640. [Dividing by 2 because garlands can be easily flipped implying that clockwise and counter clockwise arrangements are equivalent] A B C

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