Permutations & Combinations - Nikhil Goyal - Learn Quest - Part (2/2)
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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
Now what's the use of factorial in PNC ?
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!
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
simple logic is If few letters/ digits repeat
Then (Total number of letters) ! / (Letters repeats number of times )!
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!)
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
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)
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
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
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.
We have to go through the words that begin with A , then all those that begin with H and so on so forth .
Words begin with
A _ _ _ _ ( 4! = 24 words )
With O ,H ,N
I.e 24 * 4
When the first letter is R
R _ _ _ _
Now R is fixed
According to dictionary
Second letter would be A
So words with R A
R A _ _ _ (3! = 6 words )
RH (6 words )
R N( 6 words )
RO is fixed
Next again third letter will start from A
I.e ROA _ _ ( 2! = 2 words )
Next would be
Luckily we got the 4 letter as our desired one
So ROHAN ( 1 word)
rank = 24 * 4+ 6 * 3 + 2 + 1
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
R is fixed
R_ _ _ _ _
RAE ( 3! )
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
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!
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
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
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
And 8 consonants
Now since these vowels should come together
I am taking them as a unit
(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
But out of these 'h' repeats two times
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
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!)