**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 houseQ(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 allowedHow 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 allowedQ2. 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 allowedHow 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 allowedHow 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 allowedHow 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!)