Permutations and Combinations - Tricks & important formulas

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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.


Combinations

Combinations are basically 'selections without replacements' of distinct objects.

In selection, the order of the objects doesn't matter. Means, when we are selecting two distinct objects, say A and B and this is same as the selection of B and A.

Comparison of permutation and combination:
Let us consider a group of 4 distinct objects, A, B, C and D.
Combination (Selection of any two objects)Permutation (Arrangements of the selected objects)
AB AB, BA
AC AC, CA
AD AD, DA
BC BC, CB
BD BD, DB
CD CD, DC
6 combinations 12 permutations

In the above example, we can easily observe that each of the selection of two objects has 2! corresponding arrangements.

Similarly, in general, each of the selection of 'r' distinct objects should have 'r!' corresponding arrangements.

(Selections of 'r' distinct objects) * r = Arrangements of the selected objects.

i.e. Combinations of 'r' objects = Permutations of r objects / r!

Combination of 'r' distinct objects out of a total of 'n' distinct objects is expressed as nCr,

where n > 0, n ≥ r, r ≥ 0.

Therefore, nCr = nPr/r! = n!/((n-r)! r!)

In general, Combination of 'n' objects taken 'r' at a time, denoted by n and is defined as

n = ncr = n! / (n-r)! r!

Important properties and results of combinations.

  • The number of ways to select 'no' objects out of 'n' distinct objects is,
    nc0 =n!/0!(n-0)! = 1
  • The number of ways to select exactly one object out of n distinct objects is,
    nc0 =n!/1!(n-1)! = n
  • The number of ways to select 'all' the objects out of 'n' distinct objects is,
    nc0 =n!/n!(n-n)! = n
  • The number ways to select 'r' objects out of 'n' distinct objects is equal to the number of selections of 'n - r' objects out of 'n' objects.

i.e. nCr = nC(n - r)

Example: 6C4 = 6!/4!(6-4)! = 6!/4! 2!

6C(6 - 4) = 6C2 = 6!/2!(6-2)! = 6!/2! 4!

i.e. 6C4 = 6C2

How to find the numerical value of nCr easily?

Instead of applying the entire results of nCr, you can follow the below method for finding the numerical value of nCr.

Try to keep the value of 'r' as least as possible by applying the result nCr = nC(n - r).

Eg: Find the value of 10C8.

We know that 10C8 = 10C2

10C2 = Product of two consecutive numbers from 10 in descending order / Product of two consecutive numbers from 1 in ascending order= 10 *9/1 *2 = 45

Similarly;

15C3 = 15 * 14 * 13 / 1 * 2 *3

21C16 = 21C5 = 21 * 20 * 19 * 18 * 17 / 1 * 2 * 3 * 4 * 5

Values to remember:

The table of values of nC2 plays an important role in solving problems from combinations. So, if you want to become more professional in this area of counting, it is better to by heart following table.

2C2 = 1
3C2 = 3
4C2 = 6
5C2 = 10
6C2 = 15
7C2 = 21
8C2 = 28
9C2 = 36
10C2 = 45
11C2 = 55
12C2 = 66
13C2 = 78
14C2 = 91
15C2 = 105
16C2 = 120
17C2 =136
18C2 = 153
19C2 = 171
20C2 = 190

Additional important results in Combinations

In this area, we are dealing with some very important counting tricks under Combinations (Selections). These are the frequently tested results in exams. So show an extra interest in the following results and properties of Combinations.

Result 1:
Number of ways to select 'zero or more' objects out of 'n' distinct objects = 2n
OR
nC0 + nC1 + nC2 + ..... + nCn = 2n

Proof:

Here we have to count from the side of objects. Each of the 'n' distinct objects has two possibilities: either selected or not selected.

Hence the total number of selections (zero or more) out of the 'n' distinct objects = 2n

Example: A class test is conducted for 5 students, how many different ways a result card can be entered such that the result card contains only the details whether the student is passed or failed?

Each of the five students has exactly two possible results: PASS or FAIL.

Hence the total number of possible entries = 25 = 32

Result 2:
Number of ways to select 'at least one' objects out of 'n' distinct objects = 2n - 1
OR
nC1 + nC2 + nC3 + ..... + nCn = 2n - 1

Example:

How many different ways you can invite at least one among your five friends for your birthday party?

Each of your 5 friends has two possibilities: invited or not invited.

Hence the total number of possibilities = 25 = 32

But out of these 32 possibilities, one possibility is 'none among them is invited'.

Avoid this one possibility from the entire 32 possibilities for satisfying the condition of 'at least one'.

Therefore the required number of ways = 25 - 1 = 32 - 1 = 31

Result 3:

Number of ways to select 'r' objects out of 'n' identical objects = 1

Example:

Find the number of ways to select 3 toys out of 5 identical toys?

Here toys are identical, hence the number of ways = 1

Result 4:

Number of ways to select 'zero or more' objects out of 'n' identical objects = n + 1

Where 0 ≤ r ≤ n.

 

Example:

How many different ways can select zero or more apples out of 5 identical apples?

Here the difference in selections only depends upon the number of apples.

i.e. You can select 0, 1, 2, 3, 4 or 5 apples.

Hence there is 5 + 1 = 6 different selections are possible.

Result 5:

Number of ways to select 'zero or more' objects out of 'p + q + r' objects of which p objects are alike of one kind, q objects are alike of second kind and r objects are alike of third kind is {(p + 1) (q + 1) (r + 1)}

Example: How many different ways of selections of zero or more fruits are possible from 3 identical apples and 2 identical oranges?

Here again the number of selection is only based on the number of fruits from two different categories.

The possible selections can be tabulated in the following manner.

ApplesOranges
0 0
0 1
0 2
1 0
1 1
1 2
2 0
2 1
2 2
3 0
3 1
3 2
Hence a total of (3+ 1) (2 + 1) = 4 * 3 = 12 ways
Result 6:

Number of ways to select 'at least one' objects out of 'p + q + r' objects of which p objects are alike of one kind, q objects are alike of second kind and r objects are alike of third kind is {(p + 1) (q + 1) (r + 1)}- 1.

Example: How many different ways of selections of at least one fruits are possible from 3 identical apples and 2 identical oranges?

From the above illustrated tabulation of the selections, one section is (0, 0), i.e. No apple and no orange. Avoid this selection from the above mentioned total possible selections for satisfying the condition of 'at least one' fruit.

Hence the required number of selections = 12 - 1 = 11

Result 7:

Number of ways to select 'zero or more' objects out of 'p + q + r' objects of which p objects are alike of one kind, q objects are alike of second kind and r distinct objects is {(p + 1) (q + 1) 2r}.

Example: How many different ways can select 'zero or more' fruits from 2 identical apples, 3 identical oranges and 4 distinct fruits other than apple or orange?

Required number of selections = (2 +1) (3 + 1) 24 = 3 * 4 * 16 = 192

Result 8:

Number of ways to select 'at least one' objects out of 'p + q + r' objects of which p objects are alike of one kind, q objects are alike of second kind and r distinct objects is {(p + 1) (q + 1) 2r}- 1.

Example: How many different ways can select 'at least one' fruits from 2 identical apples, 3 identical oranges and 4 distinct fruits other than apple or orange?

Required number of selections = {(2 +1) (3 + 1) 24 } - 1 = (3 * 4 * 16) - 1 = 192 - 1 = 191

Result 9:

Number of selections of 'k' consecutive (adjacent) objects from a row of 'n' objects is (n - k + 1)

Examples:

Five persons are seated in a row, how many different ways can select 3 adjacent persons?

Solution:

 

From the diagram, it is easy to understand that the selections can be done in 3 ways.

i.e. (1, 2, 3), (2, 3, 4) or (3, 4, 5).

i.e. The number of ways to select 3 adjacent persons out of a row of 5 persons = 5 - 3 + 1 = 3

Geometrical application of combinations.

Total numbers of diagonals in a polygon having 'n' sides

Consider a polygon having 'n' sides (n vertices). We have to find the number of diagonals in this polygon. If we subtract the number of sides from all possible number of lines which can be formed by join these n vertices, then the remaining will be the number of diagonals.

 

Total number of lines which can be formed by join the 'n' vertices = combination of 'n' vertices taken two at a time =nc2

Note: Any questions (in this manner) based on lines are the concept of combinations.

 

Since line join from v1 to v2 in some lines join from v2 to v1 and here is no possibility to form a line by joining v1 to v1.

Replacement is not possible and there is no role for order.

Questions based on lines are combination.

Total number of sides on the polygon is 'n'

Number of diagonals = Total number of possible lines - Number of sides

= nc2 -n

Number of diagonals in an 'n' sided polygon = nc2 -n

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