Permutations & Combinations - Nikhil Goyal

Nikhil Goyal

Fundamental principle of counting

Multiplication principle : Suppose an event E can occur in m different ways and associated with each way of occurring of E, another event F can occur in n different ways, then the total number of occurrence of the two events in the given order is m × n .

Addition principle : If an event E can occur in m ways and another event F can occur in n ways, and suppose that both can not occur together, then E or F can occur in m + n ways.

Example :

There are 2 Distinct fruits and 3 vegetables available in the market. In how many ways you can pick either one fruit or one vegetable ?

Suppose there are two fruits named (F1,F2) and you have to pick one
You can either pick F1 or F2 i.e there are 2 ways to pick one fruit
Similarly for vegetables (named V1, V2 and V3) ,there are three ways to pick one vegetables ,either V1 ,V2 or V3
So total number of ways would be 2 + 3 = 5

There are 2 Distinct fruits and 3 vegetables available in the market. In how many ways you can pick one fruit and one vegetable ?

If you pick one fruit
Suppose F1 then you would have three options to pick one vegetables
Like
F1 V1
F1 V2
F1 V3
I.e 3 ways for fruit F1
Similarly three ways for fruit F2
F2 V1 , F2 V2 and F2 V3
So for every fruit (2) we have three ways to select one vegetables
I.e total number of ways = 2 * 3 = 6 ways

There are 5 doors and 6 Windows in a house
Q(1) In How many ways thief can enter in house ?
Q(2) In How many ways thief can enter through door and exist through windows ?
Q(3) How many ways thief can enter and exit ?

Solutions:

(I) Thief can enter in house ,either through doors or windows
So 5 + 6 = 11 ways
(ii) 5 * 6 = 30 ways
(iii) Here possibilities are
Thief can enter through door or window and similarly thief can exit through door or window
I.e (5 + 6) * (5 + 6) = 11 * 11 = 121 ways

Let's continue the same logic in arrangements

How many 5 digit numbers can be formed from digits 1,2,3,4and 5 such that repetition is not allowed ?

We have a 5 digit No.
i.e. 5 spaces is to be filled.
– – – – –
The 1st place can be filled with any of the 5 digits
I.e. in 5 ways (1,2,3,4,5)
As repetition is not allowed
The 2nd place in 4 ways
The 3rd place in 3 ways
The 4th place in 2 ways
The 5th place in 1 way
So, for these independent set of events, total no. of ways is 5 × 4 × 3 × 2 × 1 = 120
Similarly if the repetition is allowed
Then every space can be filled in 5 ways there is no constraint
So total number of ways = 5 * 5 * 5 * 5 * 5 = 5^5

How many numbers are there between 99 and 1000 having 7 in the units place?

First note that all these numbers have three digits.
7 is in the unit’s place.
The middle digit can be any one of the 10 digits from 0 to 9. The digit in hundred’s place can be any one of the 9 digits from 1 to 9.
Therefore, by the fundamental principle of counting, there are 10 × 9 = 90 numbers between 99 and 1000 having7 in the unit’s place.

How many numbers are there between 99 and 1000 having at-least one of their digits 7?

Total number of 3 digit numbers having at-least one of their digits as 7 =
(Total numbers of three digit numbers) – (Total number of 3 digit numbers in which 7 does not appear at all).

Total three digit numbers = 9 * 10 * 10
As first place can be filled in 9 ways and rest two can be filled in 10 * 10 ways

Total three digit number in which 7 does not appear at all = 8 * 9 * 9
As only 9 digits are available and first can be filled in 8 ways and so on ,so forth
= (9 × 10 × 10) – (8 × 9 × 9)
= 900 – 648 = 252.

Will solve some questions.

Case (i) There are 5 digits (1, 2, 3, 4, 5)
Case (ii) There are 5 digits (0, 1, 2, 3, 4)
In both the cases repetition is not allowed
How many 5 digit even numbers are possible

Solutions:

Case (i) digits are 1, 2, 3, 4, 5
For even numbers , it must ends with 2 and 4
Five spaces : _ _ _ _ _
I.e unit place can be filled in only two ways (2,4)
Now one digit is fixed ,we are left with 4 digits
Tenth place can be filled in 4 ways
Similarly rest place can be filled in 3 ,2 and 1
So total number of ways = 1 * 2 * 3 * 4 * 2 = 48 ways

Case (ii) digits are 0, 1, 2, 3, 4
For even numbers , it must ends with 0,2 and 4

Case (a) if it ends with zero
_ _ _ _ 0
I.e unit place can be filled in only one way (0)
Now one digit is fixed ,we are left with 4 digits
Tenth place can be filled in 4 ways
Similarly rest place can be filled in 3 ,2 and 1
So total number of ways = 1 * 2 * 3 * 4 * 1 = 24 ways

Case (b)
If it ends with Non-zero digits (2,4)
Five spaces: _ _ _ _ _
I.e unit place can be filled in only two ways (2,4)
Now one digit is fixed ,we are left with 4 digits
First place(ten thousand position) can be filled in 3 ways as zero can't come on this place
Now rest place can be filled in 3 ,2 and 1 as there is no constraint
So total number of ways = 3 * 3 * 2 * 1 * 2 = 36 ways

So total ways 36 + 24 = 60 ways

#Note : whenever such Question come (zero + non - zero one )
You will divide that Question into two sub-cases
first one when last digit is zero
Other one when last digit is Non-zero

Case(i) There are 5 digits (1,2,3,4,5)
Case (ii) There are 5 digits (0,1,2,3,4)
In both the cases repetition is not allowed
Q2. How many 5 digit numbers are divisible by 4 ?

Case (i)
Number to be divisible by 4,
last 2 digits should be divisible by 4.
Thus, we have the option as:
12 , 24 , 32 , 52
I.e last two.digits can be filled in these 4 ways only
Remaining three digits can be filled in 3 * 2 * 1 ways
= 6 * 4 = 24 ways

Case (ii)
In this case
We have the option as :
04 , 12, 20,24,32 , 40
Now dividing into two sub cases

Case(a)
Last two digits are zero
I.e 04 , 20 , 40
So last two digits are filled in 3 ways
Remaining three digits can be filled in 3 * 2 * 1 ways
So total number of ways 3 * 2 * 1 * 3 = 18 ways

Case (b) last two digits are Non-zero
12 , 24 ,32
So last two digits are filled in 3 ways
Now two digits has already fixed ,we are left with only 3 digits
First place can be filled in two ways only as zero can't come on this position
So remaining two places can be filled in 2 * 1 ways
So number of ways = 2 * 2 * 1 * 3 = 12 ways
Total number of ways = 18 + 12 = 30 ways

Case (i) There are 5 digits (1,2,3,4,5)
Case (ii) There are 5 digits (0,1,2,3,4)
In both the cases repetition is not allowed
How many 5 digit numbers are divisible by 3?

Case (i)
Numbers to be divisibleby 3 when sum of the digits is divisible by 3
Now we have to form a five digit number and since every digit can be used only once
So 12345 would definitely be a one five digit number
Sum of 1 + 2 + 3 + 4 + 5 = 15 divisible by 3
So number divisible by 3 = total 5 digit numbers = 120

Case (ii)
Since sum of digits = 1 + 2 + 3 + 4 + 0 = 10
Which is not divisible by 3
So zero numbers possible

Case (i) There are 5 digits (1,2,3,4,5)
Case (ii) There are 5 digits (0,1,2,3,4)
In both the cases repetition is not allowed
How many 5 digit numbers such that unit digit > ten's digit ?

Take any two digit number
Suppose 12
The possible arrangements are 12 or 21
I.e 2 cases are possible
In half of the Case unit digit > ten's digit Nd in rest half it is less than
Similarly
Taking any three digit number
123
Possible arrangements
123
132
231
213
312
321
As you can see in half cases i.e 3
Unit digit > ten's digit
123 , 213 , 312
Similarly in

case (i)
Total possible case = 120/2= 60

case (ii)
Total five digit arrangement are 96
Total possible case = 96/2 = 48

Case (i) There are 5 digits (1,2,3,4,5)
Case (ii) There are 5 digits (0,1,2,3,4)
In both the cases repetition is not allowed
How many 5 digit numbers such that unit digit > ten's digit > hundred digit ?

Taking any three digit number
123
Possible arrangements
123
132
231
213
312
321
As you can see in only one case out of 6 satisfies the given condition
Unit digit > ten's digit > hundred
123
Similarly in

case (i)
Total possible case = 120/6 = 20

case (ii)
Total five digit arrangement are 96
Total possible case = 96/6 = 16

How many 5 digit number formed by using digits (1,2,3,4,5) where 1 comes somewhere between 2 and 3 ?

Take any 5 digit number formed
Suppose 54213
Now
We have to look for the cases where
1 comes between 2 and 3
So fixing 5 and 4
54 213
54 123
54 312
54 132
54 231
54 321
As you can see
Out of these 6 arrangements , 2 are satisfying the above given condition
213 and 312
I.e 2 in every 6
I.e 1 in every 3
Here total 5 digit numbers would be 5 * 4 * 3 * 2 * 1 = 120
So 1/3 of 120 = 40

Factorial and its application in P & C

Factorial of n is N! = N * (n-1) * (n-2) * (n-3) * ... 3 * 2 * 1

And by definition
0! = 1
1! = 1
2! = 2 * 1 = 2
3! = 3 * 2 = 6
4! = 4 * 3 * 2 * 1 = 24
5! = 5 * 4 * 3 * 2 * 1 = 120
And so on

Let's see with an example

If I say how many 5 digit number can be formed by digits (1,2,3,4,5) if the repetition is not allowed
So we have 5 places
First place can be filled in 5 ways
Second in 4 ways
And so on
We would get 5 * 4 * 3 * 2 * 1
Which is basically same as 5!

So we can say that
5 distinct digits , 5 distinct positions
Total number of numbers/arrangements = 5!

Similarly
N distinct digits/alphabets, n distinct positions
Total number of arrangements = n!

How many words can be formed from the letter of RAMESH ?

Since RAMESH contain 6 different letters
So total number of words = 6! = 720

Now What if Number of letters /digits repeats and still we have to find out total number of words /numbers

Let's take an example

How many words can be formed with the letters of word Ritika ?

As you can see There are 6 letters where except 'I' every letter repeats one times and 'I' two times
Had no letter been repeated (i.e all letters are distinct )
Total number of words = 6!
But since 'I' is repeating two times
So divide 6! by 2! Ways

#Note
simple logic is If few letters/ digits repeat
Then (Total number of letters) ! / (Letters repeats number of times )!
Example :
Malayalam
M - 2 times
A - 4 times
L - 2 times
Y - 1 times
So total words = 9!/ (2! * 4! * 2!)

How many words can be formed from the letters of the word 'Kattappa' such that they don't start with K ?

Total words = 8!/ ( 2! * 3! * 2!)
And words start with K
K _ _ _ _ _ _ _
Now remaining 7 letters can be arranged in 7!/(2! * 3! * 2!) Ways
So total words don't start with K
8!/( 2! * 3! * 2!) - 7!/(2! * 3! * 2!)
=> 7 * 7!/(2! * 3! * 2!)
Or
You can directly do in this way
First position can be filled in seven ways other than K
And remaining 7 position can be arranged in
7!/(2! * 3! * 2!) Ways
So 7 * 7! / (2! * 3! * 2!)

The number of arrangements that can be made using all the letters of the word QUARTZ which begin with A but do not end with R .

QUARTZ is a six letter word
A is already fixed
A_ _ _ _ _
Last postion can be filled by 4 letters
And remaining 4 positions can be filled in 4! Ways as 4 distinct letters are there
So total number of ways 4 * 4! = 96

Sum of all numbers formed from given digits (CAT favourite)

We know n Distinct digits ,n-digit numbers , we get N! Numbers
Now let's take an example and understand the concept

Find the sum of all the four digit numbers formed using the digits 1, 2, 3 and 4 without repetition
We know 4 distinct digits , so 4! = 24 total 4 digit number would be formed
Now we have to find sum of all these 24 4-digit numbers

Case (I)
When 1 comes at thousands place in a particular number , it's contribution to the total will be 1000.
The number of numbers can be formed with 1 in the thousands place is 3! ( As rest three positions can be filled in 3! Ways )
Hence, when 1 is in the thousands place , it's contribution to the sum is 3! * 1000
Similarly when 2 comes at the thousands place
It's contribution to the sum 3! * 2000
For 3 => 3! * 3000
For 4 => 3! * 4000
I.e total sum
3! * 1000( 1+2+3+4)

Case (II)
When 1 comes in the hundreds place in a particular number , it's contribution to the total will be 100
And there are 3! Such numbers with 1 in the hundreds place.
So the contribution 1 makes to the sum when it comes in the hundreds places is 3! * 100
Similarly for 2 => 3!*200
3 => 3! * 300
And for 4 => 3! * 400
I.e total sum => 3! * (1+2+3+4) * 100

Case (III)
when 1,2,3,4 comes at the tens place
So 3! * (1+2+3+4) *10

Case (IV) when 1,2,3,4 comes at the unit place
So 3! * (1+2+3+4) * 1
So final sum
=> 3! * (1+2+3+4) * (1000+100+10+1)
=> 3! * (1+2+3+4) * (1111)

Now we can generalise

If n-digit numbers using n distinct digits are formed the sum of all the numbers so formed is equal to
(N-1)! * ( Sum of all the n digits ) * (111... N times )

Find the sum of all the numbers that can be formed by taking all the digits at a time from 1, 2, 3, 4, 5, 6 and 7 with repetition

We fixed one digit and then we arrange the remaining digit
Here since the repetition is allowed
Number can be arranged in
7 * 7 * 7 * 7 * 7 * 7 = 7^6 ways
So
7^6 ( 1+2+3+4+5+6+7) * 1111111
7^6 * (28) * 1111111

Finding the rank of a given word is basically finding out the position of the word when all the possible words have been formed using all the letters of this word exactly once and arranged in alphabetical order as in the case of dictionary .

Let's take an example

Suppose we have to find the rank of word 'ROHAN '
The letter involved here A H N O R
To arrive at the word ROHAN.
Initially
We have to go through the words that begin with A , then all those that begin with H and so on so forth .
Now
Case (i)
Words begin with
A _ _ _ _ ( 4! = 24 words )
Similarly
With O ,H ,N
I.e 24 * 4
Now
When the first letter is R
Case (ii)
R _ _ _ _
Now R is fixed
According to dictionary
Second letter would be A
So words with R A
R A _ _ _ (3! = 6 words )
Similarly
RH (6 words )
R N( 6 words )
Now
Case (iii)
RO is fixed
Next again third letter will start from A
I.e ROA _ _ ( 2! = 2 words )
Now
Next would be
ROH
Then
Case (IV)
ROHA
Luckily we got the 4 letter as our desired one
So ROHAN ( 1 word)
rank = 24 * 4+ 6 * 3 + 2 + 1
=> 117

Find the rank of word 'RAMESH'

RAMESH has letters A E H M R S
Words starting with A --> 5!
Same for E , H , M
So total words 5! *4 = 480
Now
R is fixed
R_ _ _ _ _
Case (ii)
RA
Case (iii)
RAE ( 3! )
RAH (3!)
RAM fixed
Case (IV)
RAMEHS (1 word)
RAMESH ( 1 word )
480 + 12 + 2 = 494

Combinations / Selections

Let's understand this with an example

Suppose there are three different fruits available in the market, A, B and C
You have to pick two out of three
Case (I) you can pick A and B
Case (ii) you can pick B and C
Case (iii) you can pick A and C
I.e only three cases are possible
Which can be directly find out by the formula which is NcR
Selection of R things out of N
I.e here selection of 2 fruits out of 3
Which can be done in 3c2 ways
And NcR = N! / ( N-R)! * R!
So 3c2 = 3 ways

Permutation

We don't need to learn Permutation (NpR) separately coz NpR is nothing but basically arranging those things which you have selected.
I.e NpR = NcR * R!
Example
Suppose there are three persons A,B and C and we have to select two people
We can do it in easily 3 ways
But if i have to arrange those two people
A, B can be arranged in AB and BA ways
So 3 * 2! = 6 ways
That we can directly do Without using 3p2

Although I am telling you - NpR = N!/(N-R)!

How many words can be formed using all the letters of the word PROBLEM without repetition such that vowels occupy the even places ?

Problem word has two vowels
O and E
And
There are seven places : _ _ _ _ _ _ _
On these seven places
Even places are 2,4 and 6th one
I.e for two vowels ,three positions are available
So two positions can be selected in 3c2 ways
And now these vowels can also arranged together
So 3c2 * 2! Ways
Now consonants
5 different consonants ,5 different positions
So 5! Ways
Total number of ways = 3c2 * 2! * 5!
=> 3 * 2 * 120 = 720

In how many ways can the letters of the word 'DhinchakPooja' be arranged such that vowels are always together

Dhinchakpooja has I, A, A, O, O
Five vowels
And 8 consonants
Now since these vowels should come together
I am taking them as a unit
I.e
(IAAOO) as they will always appear together
Suppose it is named as #
Now this # and 8 consonants can be arranged together
I.e total 9 letters
So 9!
But out of these 'h' repeats two times
So 9!/2!
And now vowels can also be arranged in 5!/ (2!*2!)
So total number of ways =
9!/2! * 5!/(2!*2!)

In how many ways can the letters of the word 'DhinchakPooja' be arranged such that no two vowels are together ?

Vowels can not come together
So we first fix consonants
8 consonants ,so 8 positions for consonants : _ _ _ _ _ _ _ _
Now if you carefully look
There are 9 positions available for vowels ( tip hamesha ek jyada hoti hai 8+1 = 9)
v_ v _ v _ v _ v _ v _ v_ v_ v
But vowels are only 5
So 9c5 to select 5 positions
Now these vowels can be arranged
In 5!/(2!2!)
So total number of vowels = 9c5 * 5!/(2! * 2!)
And now consonants
These can be arranged in 8!/(2!)
So total number of ways
9c5 * 5!/(2!*2!) * 8!/(2!)