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+r1Cr1
=4+31c31
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! * (10^41)/9
= (2 + 3 + 4 + 5) * 3! * (1111)Direct Formula :
Sum numbers formed by n non zero digits = (sum of digits) * (n1!) * (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+31c31=4c2
x+y+z=3
3+31c31=5c2
x+y+z=1
so 1+31c31=3c2
so total number of positive solutions will be
=4c2 * 5c2 * 3c2=180
so ordered number of positive solutions will be 180now 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 86416 * 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 =400now 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=40022/6
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