Tips on Solving Aptitude Type Questions on Permutation

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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:

1.    Formation of a team from a number of players.

2.    Formation of a particular committee from a number of players.

Examples of problems relating to permutation:

1.    Arrangement of books on a shelf.

2.    Formation of numbers with the given digits.

3.    Formation of words with the given letters.

Note: Let’s be honest. There aren’t any shortcuts here. The only way to solving these types of questions with speed and ease is by understanding the topic thoroughly. However, since Permutations and Combinations are related topics, let us first revisit them both before proceeding with tips for problems on Permutations.

This shows that while considering the alternatives of things or acts, we come across 2 types of problems: (a) Selection, (b) Arrangement.

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.

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.

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: ⁿP_r or P(n, r) = n! /(n – r)!

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

Note: Each of the different arrangements which can be made by taking some or all of a number of things at a time is called a permutation. When we have certain restrictions imposed on the arrangement or permutations of the things, we call it restricted permutations. Based on the type of restrictions imposed, these can be classified into 4 types.

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.

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.

Note: If n digits are given, including 0 and the digit at the first position is not specified, remember that the digit at the first position cannot be 0, so that the first position can have (n – 1) digits.

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! / (3! 3! 3! 3!) = 369600                                     

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.

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 n^r .

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.

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:

1.    No of circular arrangements with n elements = No of linear arrangements with n elements/n

= n! / n = (n – 1)!

2.    The number of ways in which n objects taken r at a time can form a ring = ⁿP_r / r

3.    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]

Note: So far we have been arranging the things in a straight line. However, at times we are required to arrange the objects in the form of a circle. The method of arranging things in a circle is called circular permutation.

 

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 TIPS on cracking Aptitude Questions on Combinations

Important Formulae:

The number of ways of choosing r items from a total of n items, where the order of the chosen items does not matter is denoted by ⁿC_r or C(n, r)

ⁿC_r or C(n, r) =         P(n, r)/r = n!/r!(n-r)!

[Dividing P(n, r) by r! since the order of the chosen r items does not matter]

1.    C(n, 0)= C(n, n)= 1 [We can choose all available items or none of the items in just one way]

2.    C(n, r)= C(n, n – r) [Choosing r items to be included = choosing (n-r) items for exclusion]

3.    If C(n, x)= C(n, y), then either x= y or x + y = n

4.    C(n, r) + C(n, r – 1) = C(n + 1, r) [Suppose there are n+1 items available and r items need to be selected. The n+1 items can be grouped into two sets: set A with n items and set B with just 1 item. Number of ways of choosing r items from n + 1 items = Number of ways of choosing r items when 1 item in set B is always excluded + Number of ways of choosing r items when 1 item in set B is always included]

Question: If C(n, r)= 84, C(n, r – 1)= 36 and C(n, r+1)= 126, find the value of r.

Solution:

C(n, r)/C(n, r-1) = (n – r + 1)/r = 84/36 = 7/3                                             

Again, C(n, r + 1)/C(n,r) = (n – r)/(r+1) = 126/84 = 3/2

Solving the two equations gives r = 3.

Note: Each of the different groups or selections that can be made by taking some or all of a number of things at a time, irrespective of the order, is called a combination.

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Tip #1: Selecting One or More Items

1.    When each item is unique:

Suppose there are n items and you can choose any number of items. This implies that for each item, there are 2 choices – choose that item or reject it. The total number of choices = 2 * 2 * 2 … n times = 2ⁿ

Total number of ways of choosing any number of items from a set of n items

= ⁿC₀ + ⁿC₁ + ⁿC₂ + ⁿC₃ + …. + ⁿC_n  = 2ⁿ 

Question: There are 7 questions in a question paper. In how many ways can a boy solve one or more questions?

Solution:

Number of ways in which at least one question can be selected = 2⁷ – 1= 127

[Subtracting 1 because the choice of selecting zero questions is not available]

2.    When items are not unique:

Let there be n things out of which p items are alike of one kind, q items are alike of a second kind, r items are alike of a third kind, and so on. For all p items of the same kind, we can choose 0, 1, 2... Or p items (Number of choices is p+1).

Total number of ways of selecting any number of items = (p+1)(q+1)(r+1)….

Question: In how many ways can you choose the letters of the sentence ‘Daddy did a deadly deed’? You can also select no letters at all.

Solution:

There are 9 D’s, 3 A’s, 3 E’s, 2 Y’s, 1 I and 1 L.

Total number of selections= (9+1)(3+1)(3+1)(2+1)(1+1)(1+1) = 1920

Tip #2: Dividing items into Groups

Suppose we have p + q + r items that are to be divided into 3 groups, first of size p, second of size q and the third of size r.

Number of arrangements of all items = (p + q + r)!

However, the order within each group does not matter.

The number of ways in which p + q + r items can be divided into groups containing p, q and r items respectively =    (p + q + r)!/ p! q! r!

Question: In how many ways can 52 playing cards be distributed to 4 players, giving 13 cards to each?

Solution:

The no. of ways= 52! / (13!)⁴.

Tip #3: Simultaneous Permutations & Combinations

Question: How many different words, each containing 2 vowels and 3 consonants, can be formed using all the vowels and 17 consonants?

Solution:

There are 5 vowels and 17 consonants in all.

2 vowels can be chosen in ⁵C₂ ways and 3 consonants can be chosen in ¹⁷C₃ ways.

Thus, the letters can be selected in ⁵C₂ x ¹⁷C₃ ways.

Now, each group of 5 words can be arranged in 5! Ways.

Hence, total no. of words= ⁵C₂ x ¹⁷C₃ x 5! = 816000.

Question: Out of 3 books on Economics, 4 on Political Science and 5 books on Geography, how many collections can be made if each collection consists of at least one book on each subject?

Solution:

No. of ways of choosing books on Economics= 2³ – 1= 7.

No. of ways of choosing books on Political Sciences= 2⁴ – 1= 15.

No. of ways of choosing books on Geography= 2⁵ – 1= 31.

Therefore, total number of collections that can be formed= 7 x 15 x 31= 3255.

Note: At times, the questions require us to select some things from a collection and then arrange them. These problems cannot be solved by Permutation or Combination only, but require us to apply both the methods, i.e., to apply combination to select the things and then applying permutation to arrange them in the required manner.

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