Permutations and Combinations - Tricks & important formulas

    0 Votes

Permutation and Combination is one of the most important chapters for any competitive examinations like Placement Test, Bank PO, CAT, CMAT, XAT, SNAP, FMS, IIFT, MICA, GRE and GMAT. The questions from this topic are mainly checking the skill of an aspirant in logical counting. A thorough theoretical understanding about various counting techniques will help you to tackle all the aptitude questions from this area in an interesting way with the help of tricks & important formulas.

Please go through the counting techniques article for better understanding of the basic counting skills.

It is not easy to tackle the questions under these concepts with your basic counting skills. So you should learn the mathematical applications in this regard. Here, we are mainly considering the various applications of the concept of Permutation. This is again the concept of arrangement, but only a major difference, the repetition of the objects in the arrangements is not allowable. i.e When you are facing such a situation of arrangement without repetition of the objects, understand, this is the application of Permutation.

The main objectives under this module are;
  • Factorials
  • Properties of factorials
  • Permutations(Non-circular)
  • Conditional Permutations
  • Sum of the numbers formed by different arrangements.
  • Rank of a word (or arrangement)
  • Circular Permutation
  • Combinations
  • Important properties and results of Combinations
  • Geometrical applications of Combination
  • Distribution of identical things into groups

 

Factorial

Let us consider a practical example for understanding the importance of the concept of factorial in arrangements.

Example: How many different ways can arrange the letters of the word 'FACE'?

This is basically an arrangement context, but it is an arrangement without the repletion of any letter.

There are four different letters F A, C and E.

1st letter2nd letter3rd letter4th letter
Any one among the four Anyone other than the first letter Anyone other than the first two letters Remaining one letter
4 ways 3 ways 2 ways 1 way

Therefore the total number of arrangements = 4 * 3 * 2 * 1 = 24

i.e. The product of first four natural numbers.

This is a frequently facing situation in the area of arrangements.

Here we can see the relevancy of the concept of factorial.

The product of first 4 natural numbers can be expressed as 4! (4 factorial).

i.e. n! means the product of first n natural numbers.

n! = 1 * 2 * 3 * ..... * (n - 2) * (n - 1) * n

The number of ways to arrange 'n' distinct objects in 'n' different places can be expressed as 'n!'

Hence, the number of ways to arrange one object in one place = 1! = 1 way

And the number ways to arrange '0' object (no object) in '0' place (no place) can be considered as '0!', i.e. it can be assumed in only one way.

Hence 0! = 1

"0! = 1
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5040
And so on."

Properties of factorials:

  • Factorial of any negative quantity is not valid.
  • n! = n * (n - 1) !

[i.e. 6! = 6 * 5! = 6 * 5 * 4! and so on.]

  • n!/n = (n - 1) !
  • n!/(n - 1)! = n

The numerical problems related to factorials explained in the topic 'Numbers'. Here we are considering the application of 'Factorials' in 'Arrangements'.

Here we have three different kinds of applications of factorials in regards of Non-circular Arrangements.

  • Arrangement of 'n' distinct objects.
  • Arrangement of 'n' objects in which some of them are repeated.
  • Conditional arrangements of 'n' objects (distinct or non-distinct objects).

Example for type 1:

How many different ways can arrange the letters from the word RAINBOW?

Solution:

Here we have 7 distinct letters.

Hence the required number of arrangements = 7! = 5040

Result:
The number of ways to arrange 'n' distinct letters in a row = n!

Example for type 2:

How many different ways can arrange the letters of the word INDIA?

Solution:

Here we have 5 letters and two among them are repeated.

i.e. Two 'I's are repeated. These identical letters make duplication in the number of arrangements.

For avoiding the duplication in the counting, just divide the total number of arrangements (irrespective of repetition) by 2!.

i.e. The required number of arrangements = 5! / 2! = 120/2 = 60 ways

Result:
The number of ways to arrange 'n' objects in a row such that 'p' objects out of 'n' objects are identical and 'q' objects out of 'n' objects are identical is n!/(p! q!).

Example for type 3:

How many different ways can re-arrange the letters from the word SONG so that the word should start with a vowel?

Solution:

Out of the four distinct letters, only one letter (O) is a vowel and the arrangement should start with 'O'.

i.e. The first letter can be arranged in only one way and the rest three places can be arranged in 3! ways.

Hence the total number of arrangements = 1 * 3! = 6 ways.

Permutation(Non-Circular)

Permutation means the arrangement without repetition of distinct objects. There are two different types of permutations.

  • Non-Circular Permutation.
  • Circular Permutation.

In this first part, we are considering non circular permutations. Mainly this is applicable in the context of row wise arrangements.

Result:
The number of ways to arrange 'r' objects out of 'n' distinct objects is expressed as nPr.
nPr = n!/(n-r)!, where n > 0, r ≥ 0 and n ≥ r.

Example 1: How many different three digits numbers can be formed such that the digits are distinct prime numbers?

Single digit prime numbers are: 2, 3, 5 and 7.

i.e. There are 4 distinct single digit prime numbers.

For finding the required three digit numbers, we have to arrange any three out of the above four distinct prime numbers in a row.

This is an arrangement of three out of four digits in a row.

i.e. 4P3 = 4!/ (4 - 3)! = 24

Therefore, there are 24 such distinct three digit numbers are possible.

Note: In this arrangement:

Total number of objects = 4 (four distinct single digit prime numbers)

The number of places to be filled = 3 (three digit numbers)

As per the above example, in the result nPr;
n → number of objects
r → number of places

Example 2: How many different ways 4 cars can be parked in 5 different parking slots?

Here the 4 cars are the four distinct objects and 5 available slots are the places.

Required number of arrangements = 5P4 = 5!/(5 - 4)! = 120 ways

As per the above example, in the result nPr;
n → number of objects
r → number of places

Important Note:

As per the above two examples, we can see that in a result of nPr, n and r can be either objects or places depends upon the situations of counting.

np0 = n!/(n-0)! = 1
np1 = n!/(n-1)! = n(n-1)!/(n-1)! = n
npn = n!/(n-n)! = n!/1 = n!
There for, 0!=1

Sum of the numbers formed by different arrangements.

This is an extended application of Liner permutation. All the B school entrance exams are frequently asking questions from this area. We arealready familiar withthe method of framing different numbers by the arrangements of different digits which are given. Here we are considering the method for finding the sum of all numbers formed by such arrangements.

Example:

Find the sum of all the three digit numbers formed by the digits 1, 2, and 3 without repetition.

Solution:

Without repetition of the digits we can frame 3! numbers in total, i.e. 6 numbers.

The possible numbers are given below.

123

132

213

231

312

321

Column 1Column 2Column 3
1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1
Total = 12 Total = 12 Total = 12

From the above tabulation, it is easy to understand that, after the mentioned arrangements, each column has the given digits in equal frequency but different order. This frequency or repetition of a digit in each column is equal to (n-1)!, where n is the number of distinct digits in the arrangements. In the given question we have to arrange three digits, therefore n = 3. So (n-1)! = (3 - 1)! = 2

Therefore the individual sums of each column are equal and the sum of each column can be found in the following way.

Sum in each column = sum of distinct digits in arrangement * (n - 1)!

= (1 + 2 + 3) * (3 - 1)!

= 6 * 2 = 12

Sum of all numbers = 1200 + 120 + 12 (consider the place values)

= 1332

For finding the final sum, it is possible to apply another method, i.e. 12 * 111 (number of 1's is equal to the number of places)

As per the above explanation, we can conclude the method in the following manner.

Sum of all the numbers formed by the arrangements of 'n' non zero digits
= (n - 1)! * Sum of all digits in the arrangement * 1111...n times.

Rank of a word

The rank of word means, when arranging the words (codes) formed by a certain group of distinct letters in the standard order (alphabetical order) of a dictionary, the position of a particular word (code).

Example:

If the letters from the word 'FACE' is re-arranged and the different words formed by these re-arrangements are arranged in the alphabetical order (or the order of a standard dictionary), find the rank of the word 'FACE'.

Solution:

FACE is a 4 letter word and all the words are distinct.

Alphabetical order of the letters from the word: A, C, E, F

Avoid all the words start with 'A'.

Number of words starts with A = 3! {First letter is 'A" and rest letters can be arranged in 3! ways}

Similarly;

Number of letters starts with 'C' = 3!

Number of words start with 'E' = 3!

i.e. Our required come after all of the above number of arrangements.

Consider the word start with F.

As per the alphabetical order, the first word which starts with 'F' is 'FACE', our required word.

Hence the Rank of the Word 'FACE' = 3! + 3! + 3! + 1 = 19

i.e. The 19th word of the arrangement is 'FACE'.

Example:

Find the rank of the word 'BOOK' while arranging the words formed by re-arranging the letters from the word 'BOOK' in the order of a standard dictionary.

Solution:

Letters in alphabetical order: B, K, O, O

Number of words start with 'BK' = 2!/ 2! = 1 {the letter 'O' is repeated in twice}

Number of words start with 'BOK' = 1 {Remaining letter is 'O' only}

The immediately next word is 'BOOK'.

Hence the rank of the word 'BOOK' = 1 + 1 + 1 = 3

Now we will have a look at 'Circular Permutation' and 'Combinations'. This module will give you a clear idea about the various applications of Permutations and Combinations in various practical situations, even in the area of Geometry too.

Circular permutation

Case 1: Order of the arrangement is a matter.

If the objects are arranged in a circular manner, the permutation thus formed is called circular permutation.

Total number of circular permutations of 'n' objects, ifthe order of the circular arrangement (clockwise or anti-clockwise) is considerable, is defined as (n-1)!.

Example:

The number ways to arrange 3 persons around a table = (3 - 1)! = 2 ways

The possible ways of arrangements are given below.

 

Here we are considering the arrangements in clockwise direction. Hence there are two distinct arrangements are possible and the direction of the arrangements makes a difference in the counting.

Case 2: Order of the arrangement doesn't matter.

Total number of circular permutations of 'n' objects, ifthere is 'No Difference' between clockwise and anticlockwise arrangements, is defined as (n-1)!/2.

For eg: arrangements of beads in a necklace.

In this case clockwise arrangement of the beads is not differ from the clockwise or anti clockwise arrangementsof the beads, because one among them is simply a rear view of the other. Therefore considering the order of beads in clock wise and anti clock wise are no two distinct arrangements.

We can generalize the concept of circular permutation. If we are dealing with unique objects, then the circular permutation = (n-1)!/2

Example:

The number of ways to arrange 3 different coloured beads in a necklace = (3-1)!/2 = 1

Refer the following illustration:

 

By observation, we can easily see that the second figure is simply a reflection of the first figure, if the order doesn't matter.

Hence there is only one arrangement.

i.e. The number of ways to arrange 3 different coloured beads in a necklace = 1 way.

Page 1 of 2

Popular Videos

communication

How to improve your Interview, Salary Negotiation, Communication & Presentation Skills.

Got a tip or Question?
Let us know

Related Articles

Numbers – Aptitude Test Questions, Tricks & Shortcuts
Boats and Streams - Aptitude Test Tricks, Formulas & Shortcuts
Mixtures & Alligations - Aptitude Test Tricks, Formulas & Concepts
Pipes and Cisterns - Aptitude Test Tricks, Formulas & Concepts
Time and Distance - Aptitude Test Tricks, Shortcuts & Formulas
Cyclicity of Numbers Aptitude Test Questions - Concepts Formulas Tips
Percentages - Aptitude Test Tricks & Shortcuts & Formulas
Time and Work - Aptitude Test Tricks & Shortcuts & Formulas
Problems on Trains - Aptitude Tricks & Shortcuts & Formulas
Number System Tutorial I: Integers, Fractions, Prime & Composite
Number System Tutorial II: Divisibility, Remainder, HCF & LCM
Number System Tutorial Part III: Factors, Multiples, Unit Digit and Last Two Digits of Exponents
Time & Distance Tutorial I: Basic Concepts, Average Speed & Variation of Parameters
Time and Distance Tutorial II: Relative Speed, Linear & Circular Races, Meeting Points
Time and Distance Tutorial III: Boats & Streams, Escalators, Journey with Stoppages
Logarithm - Aptitude Questions, Rules, Formulas & Tricks
Probability - Aptitude Tricks, Formulas & Shortcuts
Sequence and Series - Important concepts, Formulas & Tricks
Ratio and Proportion - Important formulas, Shortcuts & Tricks
Quadratic Equations - Aptitude Tricks