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!

    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!)


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