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


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    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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