Following types of questions pretty common in competitive exams:

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

Whether it is digits, letters or people. Approach is same. Youāve to āarrangeā 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 āorderā matters. = These are āpermutationā problems. And we can solve it without tears and without formulas. (Provided that youāve mastered my first article on PnC)
This new article contains a few more concepts related to PnC.

# 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 Aābad, Jethalal can go in 3 ways. (Because there are three train, he can pick anyone)
• Similarly From Aābad to Kutch, he can go in 7 ways.
• Pick first train from Mumbai to Aābad AND pick second train from Aābad to Kutch
• āANDā means multiplication.
• Total ways: simple multiplication =3 ways x 7 ways =21 ways Jethalal can reach Kutch.
• This is fundamental counting principle.

Ok, weāve five digits and weāve to form numbers containing five digits.
Same as 5 seats 5 men. How many ways to arrange?

Letā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 Jethalalā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

āAndā means multiplication
=5 x 4 x 3 x 2 x 1
=120 ways.

## Formula-approach

Total digits(n)=1,2,3,4,5
Weā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 doesnāt equal to 11111. Order matters!

## First chair

How many ways can you select one guy from 5?
= 5 ways. (Recall the Jethalalā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

Weāve 5 candidates left and weāve to pick one of them for second chair.
How many ways can you select one guy from 5?
= 5 ways. (Recall the Jethalalā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 youā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

āAndā means multiplication
=5x5x5x5x5
=55
=3125 ways.

# Formula: repetition allowed

Permutation of n different things taken ārā at a time, and each item is allowed to be repeated for any number of times.
In such case permutations=nr

How many numbers of five digits can be formed with the digits 1,2,3,4 and 5: with repetition?
Total items (n)=5
Weāve to take 5 digits for form a number so ārā=5
Permutation with repetitions
=nr
=55
=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 āorderā 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 weā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 Jethalalās train example)
Alternatively 4choose1=4c1=4 ways.

## How many candidates left?

Well there were four eligible candidates for the 1st 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

Weā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 ā5 digit ruleā.) 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

āAndā 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 āzeroā, weā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?

Iām only doing the last step.
=4 x 5 x 5 x 5 x 5
=4 x 54
=2500 ways.

# Case: Even numbers (without repetition)

1. How many even numbers of five digits can be formed with the digits 1,2,3,4,5 : without repetition? OR
2. How many five digit numbers can be formed using 1,2,3,4,5 which are divisible by ā2ā. (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 āEvenā number.

In our case, weā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 āANDā

## Imagine this episode

There are 5 guys wanting to sit in 5 seats.
Between Ā Mr.Two and Mr.Four, I pick one guy (2 ways) and ask him to āget-outā of the room.
So Iā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 usingĀ  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 5th (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 āGet-outā guy from Mr.Two and Mr.Four) AND (4 men in 4 seats) AND (Make that āGet-outā guy sit in fifth seat)
And means multiplication
=2 x (24) x1
=48 ways.

Formula-wise speaking
=2C1 x 4P4 x 1C1
=48 ways.

# Approach #2: The āORā 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 weāve got only 2,4.
Break the case. When we break the case, use word āORā
total possible cases
=(5 digits with ā2ā as last digit) OR (5 digits with ā4ā as last digit.)

## Broken case#1: Ā 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 1st chair to 4th chair, weā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 ā4ā as last digit

= 24 ways (just like broken case #1)
Now, combine these two broken cases.

## Total possible cases

=(5 digits with ā2ā as last digit) OR (5 digits with ā4ā as last digit.)
=24+24 (because āORā 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 doesnā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

Weāve five guys: 1,2,3,4,5
From Mr.one, three and five, I pick up one guy and tell him to āGet-outā 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 āget-outā guy back and tell me to sit in the 5th (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

1. Word ends with a vowel. (answer and approach same as Even number case)
2. Word ends with a consonant. (answer and approach same as odd number case)
3. 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 doesnāt specify the ādigitā limit.
Means 1,2,3,4,5,12,13,14,ā¦.,123,312ā¦..,53214ā¦.all digits of any length allowed as long as there is no repetition.

How many numbers possible?
Weāve to break the case. When we break the case, we use word āORā.
Possible numbers
=1 digit OR 2 digit OR 3 digit OR 4 digit OR 5 digit.