Permutation & Combination


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Will start with a basic one

    In how many ways we arrange 6 boys and 4 girls
    a) such that all girls are together
    b) all boys are together
    c) all boys are together , all girls together
    d) no girls are together

    a) all girls are together so make a group of them and take it as a one unit
    so they could be arranged in 4! ways so now we have BBBBBB(G) so total 7 persons , number of ways to arrange those 7!
    so total ways =7! * 4

    b) all boys are together so make it as one entity so they could be arranged in 6! ways
    so total number of ways = 6! * 5!

    c) all boys are together and all girls are together
    so BBBBBBGGGG
    6!*4!
    and GGGGBBBBBB
    so 4 * 6!
    so total ways = 2 * 4! * 6!

    d) no two girls are together
    so _ B _ B _ B _ B _ B _ B _
    first arrange 6 boys so ways 6!
    now 6 boys will leave 7 spaces
    now select 4 spaces and arrange 4 girls there
    so 7c4*4!
    so total ways = 6! * 7c4 * 4!

    Five boys and five girls form a line with the boys and girls alternating . The number of ways of making the line

    GBGBGBGBGB
    so 5!*5!
    now BGBGBGBGBG
    so 5!*5!
    so total ways = 5! * 5! + 5! * 5! = 2 * (5!)^2

    NUMBER OF ARRANGEMENTS that can be made with the letters of the word " MATHEMATICS " in which all vowels come together

    there are 4 vowels A,E,A,I , number of ways to arrange them = 4!/2!
    now considering four vowels as one letter ( bcz we want them together ) we have 8 letters
    M,T,H,M,T,C,S and one letter combining the vowels
    so number of ways to arrange them = 8!/(2!)*2! ( bcz of 2 M and 2 T )
    so total ways =8!/2!*2! * 4!/2!

    Concept : HOW TO FIND RANK OF A WORD

    Find the rank of the word " SACHIN" or SACHIN appears in at which number in dictionary

    IN DICTIONARY The words at each stage are arranged in alphabetic order .. we must consider words beginning wth A,C,H,I,N,S in order
    case1- number starting with A
    now we will arrange CHINS
    in 5! ways =120
    case2-NUMBER STARTING WITH C
    same we will arrange AHINS
    so 5! ways
    case3- NUMBER STARTING WITH I
    5!
    case4- number starting with N
    5!
    case5 -
    NOW NUMBER STARTING WITH S and SACHIN will appear in this list
    SACHIN is first word in list of words beginning with S
    so 5!+5!+5!+5!+5!+1=601

    Find the rank of the word DASMESH

    First consider DSMESH, 6!/2! ways=360
    DAE (arrange remaining in 4!/2!) = 12
    DAH (4!/2!) = 12
    DAM (4!/2!) = 12
    DASE (3!) = 6
    DASH (3!) = 6
    DASMEHS
    DASMESH
    so 360+50=410

    Distribute 10 similar apple to A,B,C such that A ,B,C will not get less than 2

    A+B+C=10
    now our moto in these type of questions to convert it in natural number or whole number form
    Like A+B+C=10
    we know that min value of A , B and C is 2
    so let A=a+2 ((where a > =0)
    B=b+2(b > =0)
    C=c+2((c > =0))
    so a+2+b+2+c+2=10
    so a+b+c=4
    now our question is to distribule 4 items to 3 persons such that a,b,c could get 0,1,2,,3,4
    so number of ways=n+r-1Cr-1
    =4+3-1c3-1
    6c2

    Concept : DISTRIBUTION OF IDENTICAL ITEMS IN DIFFERENT GROUPS

    let we have 30 identical mangoes and we want to distribute it to 3 person A,B and C
    a) when anyone could get any-number of mangoes ((including zero))

    so A+B+C=30
    case1- when A=0 , B+C=30
    B=0,C=30
    B=1,C=29
    .
    .
    .
    B=30 ,C=0
    so total 31 cases when A=0
    case2- Now A=1
    so B+C=29 so there will be 30 cases
    .
    .
    same for A=2 ,3 and so on
    when A=30 then B+C=0
    then B=0 and C=0 so 1 case
    so total number of ways=1+2+3.......31
    so 31 * 32/2=16 * 31 ways to distribute those mangoes to A,B,C

    NOW DIRECT FORMULA
    a1+a2+a3.....ar=n
    whole number of solution or non integral number of solutions
    =n+r-1C(r-1))
    as in this case
    A+B+C=30
    here n=30 ,r=3
    so number of ways=n+r-1cr-1
    =30+3-1c3-1
    =32c2=
    32 * 31/2 = 16 * 31

    NOW If i put any restriction here like minimum number of mangoes given to A,B,C will be 1
    so A+B+C=30
    A=1 , B+C=29
    so B=1,C=28
    B=2,C=27
    B=3,C=26
    .
    .
    B=28, C=1
    so total =28 cases
    ,
    .
    .
    now till A=28
    B+C=2
    so B=1 , C=1
    so total ways =1+2+3.....28
    so 28 * 29/2 = 14 * 29

    NOW DIRECT FORMULA
    Natural number of ways to distribute r things to n similar things
    n-1Cr-1
    here A+B+C=30
    so 30-1c3-1 = 29c2 = 29 * 28/2 = 29 * 14

    CONCEPT

    (1) SUM OF ALL NUMBERS THAT CAN BE FORMED WITH THE DIGITS 2,3,4,5 taken all at a time .. ((means repetition is not allowed))
    (2) when repetition allowed

    Approach 1)
    The total number of numbers formed with the digits 2,3,4,5 taken all at a time = 4!=24
    to find the sum of these 24 numbers we will find sum of digits at unit, ten, hundred and thousand places in all these numbers ...
    consider the digits in the unit's places in all these numbers
    Each of the digit will occur 3! =6 times in the unit place
    So total for the digits in unit's place in all the numbers = (2 + 3 + 4 +5 ) * 6
    so sum of all numbers = (2 + 3 + 4 + 5) * 3! * ( 10^0 + 10^1 + 10^2 + 10^3))
    so ((2+3+4+5) * 3! * (10^4-1)/9
    = (2+3+4+5) * 3! * (1111)

    DIRECT FORMULA
    SUM OF numbers formed by n non zero digits =
    ((sum of digits)) * (n-1!)) * (11111....n times))
    2- total number of numbers in case of repetition=4 * 4 * 4 * 4 = 4^4
    SO EACH DIGIT will occur at 4^4/4=4^3 =64 times
    so ((sum of digits)) * 64 * (1111)) so formula =sum of digits*n^n * (1111)

    CONCEPT

    FIND SUM of all numbers greater than 10000 formed by using digits 0,2,4,6,8, no digit repeated in any number

    SUM=sum of numbers formed by using digits (0,2,4,6,8)- sum of numbers formed by using digits(2,4,6,8))
    ((0+2+4+6+8) * (5-1!) * (11111)-(2+4+6+8) * (4-1!) * (1111))

    Ordered & Unordered

    Find ordered number of positive solutions of abc=30

    method-1

    abc=540
    30=2^2 * 3^3 * 5^1
    a^x * b^y * c^z=2^2 * 3^3 * 5^1
    a will some number which will contain 2 ,3 and 5
    same for b
    and same for
    and there will be some power of 2 ,3 and 5 in them whatever will be the power 0,1,2 or 3
    but sum of the power of 2 will be 2 , sum of the power of 3 will be 3 and sum of the power of 5 will be 1
    so x + y + z = 2 (where x,y,z could vary from 0 to2)
    so number of ways=2+3-1c3-1=4c2
    x+y+z=3
    3+3-1c3-1=5c2
    x+y+z=1
    so 1+3-1c3-1=3c2
    so total number of positive solutions will be
    =4c2 * 5c2 * 3c2=180
    so ordered number of positive solutions will be 180

    now if we want total number of integral solutions then we will find negative solutions also
    for negative out of a,b,c ,2 numbers will be negative
    so choose 2 numbers out of those 3 numbers
    so 3c2* positive solutions
    so 3c2 * 180
    so total ordered solutions =180+3c2 * 180
    abc=300
    abc=2^2 * 3 * 5^2

    so x+y+z=2 so 4c2
    x+y+z=1 so 3c2
    x+y+z=2 so 4c2
    so positive ordered
    = 4c2 * 3c2 * 4c2
    6 * 6 * 3 = 108 positive ordered
    and total 108 + 3c2 * 108

    Unordered:cases when all are different
    cases when 2 same one different
    cases when all are same
    a * b * c = 2^2 * 3 * 5^2
    when 2 are same
    a^2 * c = 2^2 * 3 * 5^2
    (1)^2 * (X)
    (2)^2 * (X)
    (5)^2 * (X)
    (10)^2 * (X)

    so 4 cases will be there and way to arrange them is 3!/2!=3 cases
    and when all are same
    a^3=2^2 * 3 * 5^2
    not possible
    so 6x+y+z=108
    so x=108-12/6
    so x=16
    and y=4
    so unordered =x+y=20

    If a * b * c * d = 648, then how many ordered integral values of a,b,c,d are possible...

    648=2^3 * 3^4
    so x+y+z+k=3 so 6c3=20
    and x+y+z+k=4 =7c3=35
    so total positive =700
    now cases can be
    all positive=700
    2 positive 2 negative =4c2 * 700
    or all negative =700
    so 700+700+6 * 700
    4200+1400=5600
    a^x * b^y * c^z * d^k = 2^3 * 3^4
    now a ,b,c,d all will contain 0,1,2,or 3 power of 2 and sum of all those will be equal to 3
    so x+y+z+k=3
    and same for 3
    so x+y+z+k=4

    In how many ways can 5^17 be written as a product of three positive numbers
    a+b+c = 17
    19C2 = 171
    when two are same :
    2a+c = 17
    total 9 solutions
    (171-3 * 9)/ 3! + 9 = 33

    No need of making cases!
    x + y + z = 17
    when 2 are equal then
    2x+y=17
    so 9 cases

    or make cases
    (1)^2*x
    (5^2)*x
    (5^4)*x
    (5^6)*x
    ,.
    .
    (5^16)*x

    |x * y * z| =210.find number of Integral Solutions

    210= 2 * 3 * 5 * 7 ..
    So positive = 3 * 3 * 3 * 3 = 3^4
    Now each of three can be either positive or negative
    So 2 * 2 * 2 * 3^4 = 8 * 3^4
    Unordered:81 - 3 / 6 +1 = 14

    in how many ways can 3^11 can be expressed as a product of 3 numbers
    a) 16
    b) 27
    c) 82
    d) 10

    In how many ways 1001^2 can be written as product of 3 integers
    total ordered and unordered

    a * b * c = 7^2 * 11^2 * 13^2
    x+y+z=2=4c2=6
    x+y+z=2=4c2=6
    x+y+z=2=4c2=6
    so 6^3+3 * 6^3
    4 * 6^3=864
    now unordered
    2 same
    a^2 * b=1^2 * x
    7^2 * x
    11^2 * x
    13^2 * x
    (7 * 11)^2 * x
    (7 * 13)^2 * x
    (11 * 13)^2 * x
    (7 * 11 * 13)^2
    same 8 for negative so 16
    so 864-16 * 3/6 +16

    In how many ways 1000 can b expressed as the product of 3 integers?
    ordered and unordered

    1000=2^3 * 5^3
    so x+y+z=3
    so 5c2=10
    and x+y+z=3
    so 10
    so 10 * 10=100
    and negative 3 * 100=300
    so total =400
    now unordered
    same a^2 * b=2^3 * 5^3
    1^2 * x
    2^2 * x
    5^2 * x
    10^2 * x
    (-1)^2 * x
    (-2)^2 * x
    (-5)^2 * x
    (-10)^2 * x
    so total y=8
    but one case will be all same here so y=7
    and z=1
    so 6x + 3 * 7 + 1=400
    so x=400-22/6
    so unordered=x+y+z ... this is total number of ordered and unordered solutions


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