@Amrofa said:If x, y and z are whole numbers such that x > y, then how many solutions are possible for the equation x + y + z = 36?
171??
@Amrofa said:If x, y and z are whole numbers such that x > y, then how many solutions are possible for the equation x + y + z = 36?
okay, this is not the first time i am doing this question, the previous time i wasn't able to do it but i remember the solution from that time....
x + y + z = 36
total ways of distribution in this way= (36 +3-1)C(3-1)
= 19*37
Now from these cases subtract those with x=y and x less than y
19 cases.... x=y=0 to x=y=18
remaining cases = 19*36
now of these cases half will have x>y and half x
less than y
answer = 19*18 = 342
x + y + z = 36
total ways of distribution in this way= (36 +3-1)C(3-1)
= 19*37
Now from these cases subtract those with x=y and x less than y
19 cases.... x=y=0 to x=y=18
remaining cases = 19*36
now of these cases half will have x>y and half x
less than y
answer = 19*18 = 342
@Amrofa said:If x, y and z are whole numbers such that x > y, then how many solutions are possible for the equation x + y + z = 36?
Total no. of solutions=38C2
cases where x=y : 19
cases where x>y : (38C2-19)/2=684/2=342
@Amrofa said:If x, y and z are whole numbers such that x > y, then how many solutions are possible for the equation x + y + z = 36?
342
@Amrofa said:If x, y and z are whole numbers such that x > y, then how many solutions are possible for the equation x + y + z = 36?
@Zedai said:190?
@x2maverickc said:total no of possible solutions=38C2out of which in one case x=y.so solns in which x>y will be half of (38C2 - 1). will be 351
@ravi.theja said:331??
my take - 342
x+y+z=36 has 38C2 solutions
in these (x,y) = (0,0), (1,1), ... (18,18) have to be removed.
after that half the numbers have to be removed since in them y>x
final answer (38C2-19)/2 = 342
@adityaknsit said:okay, this is not the first time i am doing this question, the previous time i wasn't able to do it but i remember the solution from that time....x + y + z = 36total ways of distribution in this way= (36 +3-1)C(3-1)= 19*37Now from these cases subtract those with x=y and x1. x=y19 cases.... x=y=0 to x=y=18 remaining cases = 19*36now of these cases half will have x>y and half xanswer = 19*18 = 342
keeping z = 1 X varies from 18 17 1 - 34 1 1 (17 cases)
keeping z = 2 X varies from 18 16 2 - 33 2 1 (16 cases)
and so on
till z = 33 where X varies from 2 1 33 (1 case)
so 1 + 2 + 3 + .... + 17 = 153 ??
keeping z = 2 X varies from 18 16 2 - 33 2 1 (16 cases)
and so on
till z = 33 where X varies from 2 1 33 (1 case)
so 1 + 2 + 3 + .... + 17 = 153 ??
@Amrofa said:If x, y and z are whole numbers such that x > y, then how many solutions are possible for the equation x + y + z = 36?
x+y+z=36 can be done in 38C2 ways = 703 ways
Out of which, x=y will be 19 times ( x and y range from 0 to 18 )
So, excluding x=y, we have 684 ways.
Out of which, half the times x>y and half the times y>x.
So, x>y => 342 times
Out of which, x=y will be 19 times ( x and y range from 0 to 18 )
So, excluding x=y, we have 684 ways.
Out of which, half the times x>y and half the times y>x.
So, x>y => 342 times
Can anyone please explain me in detail the concept behind this problem?
The least number which when divided by 35,45 and 55 leaves a remainder 3 but when divided by 9 leaves no remainder, is
A. 1683
B. 1677
C. 2523
D. 3363
Find number of solutions for a,b such that a+b+LCM(a,b)=89?
@Zedai said:@anytomdickandhary Sir, please help me out to solve this question using equations:3 different dies were rolled. find the no. of ways the summation of no will be 12?my approach:x+y+z=12(6-x')+(6-y')+(6-z')=12x'+y'+z'=6no of non negative values: 8C2=28 where am i going wrong with the concept sir?
here x,y,z are not >=0 but >=1.
@pirateiim478 said:Find number of solutions for a,b such that a+b+LCM(a,b)=89?
a=hx
b=hy
h is the hcf
so h(x+y+xy)=89
but 89 is prime.
so h has to be 1 and x+y+xy=89
(x+1)(y+1)=90=2*3^2*5
2*3*2=12 solutions
(1,90)
(2,45)
(3,30)
(5,18)
(6,15)
(9,10)
(10,9)
(15,6)
(18,5)
(30,3)
(45,2)
(90,1)
Out of these in 10 cases, x,y are coprime
so 10 solutions
b=hy
h is the hcf
so h(x+y+xy)=89
but 89 is prime.
so h has to be 1 and x+y+xy=89
(x+1)(y+1)=90=2*3^2*5
2*3*2=12 solutions
(1,90)
(2,45)
(3,30)
(5,18)
(6,15)
(9,10)
(10,9)
(15,6)
(18,5)
(30,3)
(45,2)
(90,1)
Out of these in 10 cases, x,y are coprime
so 10 solutions
@bs0409 said:a=hxb=hyh is the hcfso h(x+y+xy)=89but 89 is prime.so h has to be 1 and x+y+xy=89(x+1)(y+1)=90=2*3^2*53 solutions
bs bhai Shocked to see your cat result. May be 2nd section spoiled your score. I too got low score this time 91%le. Preparing seriously this time!! Lets rock this time (Y)
@bs0409 said:a=hxb=hyh is the hcfso h(x+y+xy)=89but 89 is prime.so h has to be 1 and x+y+xy=89(x+1)(y+1)=90=2*3^2*53 solutions
please explain the last step (x+1)(y+1)=90
is this mathematical induction that you are using ?
is this mathematical induction that you are using ?
@pirateiim478 said:bs bhai Shocked to see your cat result. May be 2nd section spoiled your score. I too got low score this time 91%le. Preparing seriously this time!! Lets rock this time (Y)
yes...2nd section=70%ile but it is not shocking since I did not study second section and ought to be punished..........:D
How many four-digit numbers are there with less than 6 different prime factors?
(1) 1224 (2) 8476 (3) 9000 (4) 7613 (5) none of these
How many four-digit numbers are there with less than 6 different prime factors?
(1) 1224 (2) 8476 (3) 9000 (4) 7613 (5) none of these
@somnathbhatta said:please explain the last step (x+1)(y+1)=90 is this mathematical induction that you are using ?
It has got nothing to do with induction.Just breaking up a number into two different factors....