Following types of questions pretty common in competitive exams:

- How many numbers of five digits can be formed with the digits 1,2,3,4 and 5, without repetition?
- How many words can be formed using all of the letters from A to E without repetition?
- How many ways can 5 men be arranged in 5 chairs?

Whether it is digits, letters or people. Approach is same. You’ve to ‘arrange’ them. But When we form numbers, 12345 is not same as 54321. Both numbers are different.Similarly ABDCE word is not same as BCDEA. So order matters, these are a “**Permutation**” problems

It means “order” matters. = These are “permutation” problems. And we can solve it without tears and without formulas. (Provided that you’ve mastered my first article on PnC)

This new article contains a few more concepts related to PnC.

- Case: Without repetition
- Case: With repetition
- Case: Entry of ZERO (without repetition)
- Case: Entry of ZERO (with repetition)
- Case: Even numbers (without repetition)
- Case: Odd numbers
- Mock Questions
- Case: All possible numbers

# Case: Without repetition

How many numbers of five digits can be formed with the digits 1,2,3,4 and 5, without repetition?

Repetition of digits, is not allowed = we cannot form numbers such as 22222, 23355 etc. Each digit must be different in this case.

Similarly AAABC or BBCAD is not allowed.

# Approach: Fundamental counting

Recall the first article on PnC

There are 3 trains from Mumbai to Ahmedabad and 7 trains from Ahmedabad to Kutch. In how many ways can Jethalal reach Kutch?

- From Mumbai to A’bad, Jethalal can go in 3 ways. (Because there are three train, he can pick anyone)
- Similarly From A’bad to Kutch, he can go in 7 ways.
- Pick first train from Mumbai to A’bad
**AND**pick second train from A’bad to Kutch - “AND” means multiplication.
- Total ways: simple multiplication =3 ways x 7 ways =21 ways Jethalal can reach Kutch.
- This is fundamental counting principle.

Ok, we’ve five digits and we’ve to form numbers containing five digits.

Same as 5 seats 5 men. How many ways to arrange?

Let’s start filling up the chair, one chair at a time.

## First chair

How many ways can you select one guy from 5?

= 5 ways. (Recall the Jethalal’s train example)

*Alternatively 5choose1=5c1=5*

How many candidates left?

One guy occupied the first chair so 5-1=4 guys left.

## Second chair

Ok now 4 guys waiting to seat in the chair.

How many ways can you select one guy from 4?=4 ways.

*Alternatively 4c1=4 ways.*

Now first chair and second chair is occupied.

Means 2 out of 5 gentlemen got the seats and 5-2=3 guys still waiting to sit in a chair.

## Third chair:

Candidates left: 3

How many ways can you select one guy from given 3? =3 ways.

## Fourth Chair

Now 2 candidates left.

How many ways can you select one guy from given 2? =2 ways.

## Fifth Chair

Only 1 candidate left.

How many ways can you select one guy from given 1? =Obviously 1 way.

In short,

## Total ways to arrange 5 digit numbers

=First chair AND second chair and third chair and fourth chair and fifth chair

“And” means multiplication

=5 x 4 x 3 x 2 x 1

=120 ways.

## Formula-approach

Total digits(n)=1,2,3,4,5

We’ve to form numbers containing 5 digits. So r=5.

Permutation

=nPr

=5P5

=5!/(5-5)!

=5!/0!

=5!/1 (because 0!=1)

=5x4x3x2x1

=120 ways.

Final answer: from the given digits 1,2,3,4,5 we can for 120 numbers which contain 5 digits.

# Case2: With repetition

How many numbers of five digits can be formed with the digits 1,2,3,4 and 5: with repetition?

This time we can repeat digits, for example 22233,11111 etc. numbers are allowed.

But it is still a permutation problem because 22233 doesn’t equal to 11111. Order matters!

## First chair

How many ways can you select one guy from 5?

= 5 ways. (Recall the Jethalal’s train example)

Alternatively 5choose1=5c1=5

## How many candidates left?

BUT, immediately after one guy is seated in first chair, **his clone** appears and stands in the waiting line.

So how many candidates left? 5 candidates.

# Second chair

We’ve 5 candidates left and we’ve to pick one of them for second chair.

How many ways can you select one guy from 5?

= 5 ways. (Recall the Jethalal’s train example)

Alternatively 5choose1=5c1=5

BUT, immediately after that guy is seated in second chair, **his clone** appears and stands in the waiting line.

So how many candidates left? 5 candidates.

## Third chair

Again 5 candidates left and you’ve to pick one guy for third seat= 5 ways.

But again his clone appears and stands in the waiting line and 5 candidates left.

Process continues till all chairs are occupied like this.

## Total numbers

=First chair AND second chair and third chair and fourth chair and fifth chair

“And” means multiplication

=5x5x5x5x5

=5^{5}

=3125 ways.

# Formula: repetition allowed

Permutation of n different things taken “r” at a time, and each item is allowed to be repeated for any number of times.

In such case permutations=n^{r}

How many numbers of five digits can be formed with the digits 1,2,3,4 and 5: with repetition?

Total items (n)=5

We’ve to take 5 digits for form a number so “r”=5

Permutation with repetitions

=n^{r}

=5^{5}

=3125

# Case: Entry of ZERO (without repetition)

How many numbers of five digits can be formed with the digits 0,1,2,3,4 : without repetition?

It is a permutation problem, because “order” matters.

But 01234 is not a five digit number, it is a four digit number.

Means Mr.zero cannot occupy the first chair, although he is eligible for any other remaining chairs.

## First chair

Number of eligible candidates = 4.

(because Mr.Zero is not allowed, so we’ve *Mr.One, Mr.Two,Mr.Three and Mr.Four* = four candidates)

How many ways can you select one guy from 4?

=4 ways. (Recall the Jethalal’s train example)

Alternatively 4choose1=4c1=4 ways.

## How many candidates left?

Well there were four eligible candidates for the 1^{st} chair and one of them occupied the first chair.

So 4-1=3 candidates left.

BUT, now Mr.Zero is also eligible for the remaining seats #2,#3,#4 and #5.

So remaining candidates =3 candidates + Mr.Zero=4 candidates.

## Second chair

We’ve four candidates left (including Mr.Zero) and each of them is eligible to compete for second chair.

How many way can you select 1 guy out of 4? = 4 ways.

(We had to be careful only for first chair because of the “5 digit rule”.) from now on, things will move smooth just like case #1.

How many candidates left?

4 guys competing for second chair. One of them got selected so 4 -1= 3 guys left.

## Third chair

How many way can you select 1 guy out of 3 guy left from previous chair? = 3 ways.

Now only 2 guys left.

Second chair= 2 ways

Now 1 guy left

First chair = 1 way.

Total numbers containing 5 digits

=First chair AND second chair and third chair and fourth chair and fifth chair

“And” means multiplication

=4 x 4 x 3 x 2 x 1

=96 ways.

## Similar question

There are 5 gentlemen in Gokuldham society Jetha, Sodhi, Bhide, Mehta and Aiyyar. How many ways can they be arranged in 5 chairs with condition that Jethalal must not occupy the first chair?

= same approach, same answer. Instead of 5 numbers we have 5 men and instead of “zero”, we’ve Jethalal.

# Case: Entry of ZERO (with repetition)

How many numbers of five digits can be formed with the digits 0,1,2,3,4 : **with repetition?**

Do this on your own.

I’m only doing the last step.

=4 x 5 x 5 x 5 x 5

=4 x 5^{4}

=2500 ways.

# Case: Even numbers (without repetition)

- How many even numbers of five digits can be formed with the digits 1,2,3,4,5 : without repetition? OR
- How many five digit numbers can be formed using 1,2,3,4,5 which are divisible by “2”. (without repetition)?

## What is an even number?

If the last digit of a given number is 0,2,4,6 or 8, we call it “Even” number.

In our case, we’ve five digits: 1,2,3,4,5

If we want to form even numbers, the last digit (fifth chair) must contain either 2 or 4.

Now we can solve this case, using two approaches

# Approach#1: The “AND”

## Imagine this episode

There are 5 guys wanting to sit in 5 seats.

Between Mr.Two and Mr.Four, I pick one guy (2 ways) and ask him to “get-out” of the room.

So I’ve 5 minus 1 = 4 guys left in the room.

I fill up first, second, third and fourth chair using these 4 guys.

How many ways can I fill up 4 chairs using 4 men?

4x3x2x1=4!=24 ways.

(alternatively, apply formula =4P4)

I had sent one guy outside, now I call him back and ask him to sit in the 5^{th} (last) chair. Since there is only one guy left and one chair left=1 way. Combine this episode into single line

# Total even number

=(pick one “Get-out” guy from Mr.Two and Mr.Four) AND (4 men in 4 seats) AND (Make that “Get-out” guy sit in fifth seat)

And means multiplication

=2 **x** (24) **x**1

=48 ways.

Formula-wise speaking

=2C1 x 4P4 x 1C1

=48 ways.

# Approach #2: The “OR” Approach (Breaking the case)

Same question:

How many even numbers of five digits can be formed with the digits 1,2,3,4,5 : without repetition?

Even number means last digit has 0,2,4,6 or 8.

In our case we’ve got only 2,4.

Break the case. When we break the case, use word “OR”

total possible cases

=(5 digits with “2” as last digit) OR (5 digits with “4” as last digit.)

## Broken case#1: Five digits number with 2 as last digit.

It means MR.2 must always occupy last chair. (=1 way)

So we had five guys but one of them Mr.two is not eligible to sit in any other chair except last one.

=5-1=4 guys left

And From 1^{st} chair to 4^{th} chair, we’ve total 4 chairs.

4 chairs and 4 guys

Permutation

=4 x 3 x 2 x 1

=24 ways.

(Alternatively 4P4=4!=24)

## broken case #2: Five digits number with “4” as last digit

= 24 ways (just like broken case #1)

Now, combine these two broken cases.

## Total possible cases

=(5 digits with “2” as last digit) OR (5 digits with “4” as last digit.)

=24+24 (because “OR” means addition (+)

=48

# Case: Odd numbers

How many Odd numbers of five digits can be formed with the digits 1,2,3,4,5 : without repetition?

What is an odd number?

- It has 1,3,5,7 or 9 as last digit
- Means it doesn’t have 0,2,4,6 or 8 as last digit.

## Shortcut

So far you know that

Total number possible (from case#1)=120

Total even numbers possible (from previous case)=48

So odd numbers = total minus even number

=120 minus 48

=72 odd numbers possible.

## Usual approach

We’ve five guys: 1,2,3,4,5

From Mr.one, three and five, I pick up one guy and tell him to “Get-out” of the room =**3 ways.**

How many guys left ?

There were five in the room, one of them went outside so 5-1=4 guys left.

Arrange them in seat number one to four = **4! Ways**.

Call that “get-out” guy back and tell me to sit in the 5^{th} (last) seat =**1 way**

## Total odd numbers

=3 ways x 4! ways x 1 way

=3 x 24

=72

# Mock Questions

Without considering dictionary meaning, How many words can be formed using all letters from A to E such that

- Word ends with a vowel. (answer and approach same as Even number case)
- Word ends with a consonant. (answer and approach same as odd number case)
- Both first and last letter of the word are vowels. (Now you think about that!)

# Case: All possible numbers

How many numbers can be formed using digits 1,2,3,4,5 : without repetition?

It doesn’t specify the “digit” limit.

Means 1,2,3,4,5,12,13,14,….,123,312…..,53214….all digits of any length allowed as long as there is no repetition.

How many numbers possible?

We’ve to break the case. When we break the case, we use word “OR”.

Possible numbers

=1 digit OR 2 digit OR 3 digit OR 4 digit OR 5 digit.

“OR” means addition

=1 digit number + 2 digit number + 3 digit number + 4 digit number +5 digit number

Number of digits in the number | Ways to arrange | ||

1 (e.g.2) | 5 men 1 seat | =5 | 5 |

2 (e.g.23) | 5 men 2 seats | =5×4 | 20 |

3 (e.g.234) | 5 men 3 seats | =5x4x3 | 60 |

4 (e.g.2345) | 5 men 4 seats | =5x4x3x2 | 120 |

5 (e.g. 23451) | 5 men 5 seats | =5x4x3x2x1 | 120 |

Total | 325 |

## Formula approach

=5P1 + 5P2 + 5P3 +5P4 +5P5

=325

**Final answer**: From the given digits 1,2,3,4,5 we can form total 325 numbers.

visit Mrunal.org/APTITUDEto read my other articles on Aptitude.

Still having Trouble solving some PnC question from your book? Post is in comments below!

## So far 13 Comments posted

Answers for mock ques above:

1. 48

2. 72

3. 6.?

Ans for mock que..

1. 48

2. 72

3. 11??

The Ans for Mock question

1. 48

2. 72

3. 12

answers to 1,2 are right..

Q3..,ITS 12.. SINCE A and E can interchange themselves..,they can arrange themselves in 2 ways..either row starts with A and ends with E.., OR IT starts with E and ends with A..,

B,C,D can arrange themselves in 3 factorial = 6..

so answer is 6 x 2 =12..

ans for 1 and 2 are right. ans for 3rd wud be 12. bcoz one can place A and E either in 1st or last position. so there are two ways to select vowels at ends…. A_ _ _ E or E_ _ _ A.

helo mrunal ur link “my first article on PnC” for basics of permutations and combimations lead to another page and bnot the intended one.. kindly do needful

thanks for notifying.

you can access the Archive of all articles, on following link

mrunal.org/aptitude

Mrunal sir its unable to acess d link of PERCENTAGE not found while assesing …IN APTITUDE COLUMN……wy dis link is not found..pls help us.OR GIVE ANOTHER LINK………ALSO PUT SOME CHAPTERS ON TOPPIC AVERAGE,NUMBER SYSTEM,TRIGO.ETC

I M FEEL VERY COMFORTABLE WITG PERMUTATION AFETR STUDY MRUNAL SIR ARTICLES…

THANKING U SIR

thanx sir….u were very helpful in clearing our concepts…..god bless you

HI Mrunal,

I am Facing a probability problem

A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

Why I am not getting the right answer by following the below approach

Suppose two drawn balls are blue then their probability out of total will be will be 2C2/7C2 which is 1/21, and when we subtract it from 1 we should get the probability of not getting these balls which is 1-1/21 that is 20/21 , but the correct answer is 10/21,

please clarify,

How to find the sum of all digits formed by ‘x’ nos. of 4 , ‘y’ nos. of 5, ‘z’ nos. of 6? There is no limitation on no. of digits.

Prem sethi

In total you have 7 balls

So you can pick them in 7*6 ways

To avoid same cases 7*6/2*1=21

I.e. 21 = total possible cases

Now you can make a talent pool carrying 2 blue balls

So left with 5 balls

Desired cases=5*4/2*1=10

Prob. =10/21