# Permutations & Combinations - Nikhil Goyal - Learn Quest - Part (1/2)

• 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

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