Quant Boosters - Hemant Malhotra - Set 11



  • 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



  • Q21) Sum of all numbers that can be formed with the digits 2, 3, 4 and 5
    a) Repetition not allowed
    b) Repetition allowed



  • 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 numbers formed by n non zero digits = (sum of digits) * (n-1!) * (11111....n times)

    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)



  • Q22) Find ordered number of positive solutions of a * b * c = 540



  • abc = 540
    540 = 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



  • Q23) 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



  • Q24) 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



  • Q25) |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



  • Q26) In how many ways 1001^2 can be written as product of 3 integers (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



  • Q27) 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



  • Q28) Find coefficient of x^3 * y^4 * z^3 in expansion of (2x + y + 4z)^10



  • general term of this equation is = 10!/(r1! * r2! * r3!) * (2x)^r1 * (y)^r2 * (4z)^r3
    = 10!/(r1! * r2! * r3!) * (2)^r1 * (4)^r3 * x^r1 * y^r2 * z^r3
    now we have to find x^3 * y^4 * z^3 so r1=3 ,r2=4 and r3=3
    put tvalue of r1, r2, r3 here 10!/(r1! * r2! * r3!) * (2)^r1 * (4)^r3 * x^r1 * y^r2 * z^r3
    so coefficient is 10!/(3! * 4! * 3!) * (2)^3 * (4)^3



  • Q29) Find the largest coefficient in (x + y + z + k)^15



  • general term = 15!/(r1! * r2! * r3! * r4!) * x^r1 * y^r2 * z^r3 * k^r4
    so basically we have to maximize 15!/(r1! * r2! * r3! * r4!) term
    so we have to minimize (r1! * r2! * r3! * r4!) so for this symmetry is useful so put values as close as possible here
    r1 = r2 = r3 = 4 and r4 = 3
    so largest coefficient is 15!/(3! * (4!)^3)



  • Q30) Find the coefficient of x^6 in (1 + x^2 - x^3)^8


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