@meow14 said:@techgeek2050@mailtoankit correct. plz post approach.
x1 + x2 + x3 + x4 = 6
number of ways 9C3 = 84
@meow14 said:@techgeek2050@mailtoankit correct. plz post approach.
4 fingers (a,b,c,d)
a + b + c + d = 6
(6 + 4 - 1)c(4 - 1) = 9c3 = 84
Find the largest value of y/x if x ˛ + y ˛ - 2x + 4y - 19 = 0.
Dont have OA please discuss .
Dont have OA please discuss .
@mailtoankit said:4 fingers (a,b,c,d)a + b + c + d = 6(6 + 4 - 1)c(4 - 1) = 9c3 = 84
can u plz expln the formula u r using? 
@meow14 said:@techgeek2050x1???
x1 - number of rings in first finger
x2 - number of rings in 2nd finger
... and so on
@Asfakul said:Find the largest value of y/x if x ˛ + y ˛ - 2x + 4y - 19 = 0.Dont have OA please discuss .
i guess it should be infinite.
@meow14 said:@techgeek2050@mailtoankit correct. plz post approach.
haww... mera bhi thik thaa.... 😞 meko tag nhi kia aapne!!!
on second thoughts acha kia nhi kia... my method is pretty time consuming.
anyway, this is how i did.
all 6 rings in 1 finger 4
for 2 fingers there are 5 cases, 5*4C2 = 30
for 3 fingers there'll be 10 cases. so.. 10 * 4C3 = 40
for 4 fingers ,...10
add all of them.. 84~
on second thoughts acha kia nhi kia... my method is pretty time consuming.
anyway, this is how i did.
all 6 rings in 1 finger 4
for 2 fingers there are 5 cases, 5*4C2 = 30
for 3 fingers there'll be 10 cases. so.. 10 * 4C3 = 40
for 4 fingers ,...10
add all of them.. 84~
center of circle is at (1, - 2)
max y = 3 (approx) , is at x = 1y / x = 3 (approx)
dy/dx = 0 at x = 1
dy/dx >0 when 0
when 0 1
so as x tends to 0, y / x increases
y/x will go on increasing as x approaches 0
m not sure though. Let @bodhi_vriksha sir post the OA
@meow14 said:@mailtoankit why a+b+c+d=6??
suppose a,b,c,d to be four fingers..it is given in question that a finger can have 0 to 6 rings possible.
so question translates to "finding the number of non-negative solution of a+b+c+d = 6"
hope that helps..
so question translates to "finding the number of non-negative solution of a+b+c+d = 6"
hope that helps..
@meow14 said:can u plz expln the formula u r using?
for distribution of "n" identical things in "r" diff. things...
no. of non - negative integral solutions = (n + r - 1)c(r - 1)
for positive integral solutions = (n - 1)c(r - 1)
where n = 6 identical rings....r = 4 diff. fingers
@meow14 said:@ChirpiBird VERY SORRY!!I also did in a similar way. But the shortcut samajhme nahin aaya.
arre it's okay! i was kidding. 
shortcut samjhaun?
it's like finding the number of non neg integral solutions to this eq.
a+b+c+d= 6
4+6-1 C 4-1
where a,b,c,d are fingers and 6 rings.
general eq is
x1 +x2 + x3 ... + x(r) = n
then non negative integral solutions to this eq are ... n+r-1Cr-1
got it??
shortcut samjhaun?
it's like finding the number of non neg integral solutions to this eq.
a+b+c+d= 6
4+6-1 C 4-1
where a,b,c,d are fingers and 6 rings.
general eq is
x1 +x2 + x3 ... + x(r) = n
then non negative integral solutions to this eq are ... n+r-1Cr-1
got it??
The number of divisors of the form 4n + 1 , n>=0 , of the number (10^10)*(11^11)*(13^13) is .... ?
approach plz.
@meow14 said:@techgeek2050@mailtoankit correct. plz post approach.
This is how we approach these questions:
this is case where we need to distribute 6 similar things between 4 different objects(S->D)
a+b+c+d=6
where a,b,c,d represent the fingers( or different objects in which we want to distribute)
now the scenario is you have 9 things(6 +3=9, 6=> is present on RHS, 3=> is the count of plus sign) out which 6 are same(treat them as 1's) and 3 are same (treat them as 0's) and you want to distribute them.
It will take care of all the cases possible.
how you pretty well know how to distribute them..
OA : 9C3=84
This approach can be followed any where where you have a situation of distributing the same things to different objects. It is called as Partition method or one-zero approach.
I hope you like this approach. :)
this is case where we need to distribute 6 similar things between 4 different objects(S->D)
a+b+c+d=6
where a,b,c,d represent the fingers( or different objects in which we want to distribute)
now the scenario is you have 9 things(6 +3=9, 6=> is present on RHS, 3=> is the count of plus sign) out which 6 are same(treat them as 1's) and 3 are same (treat them as 0's) and you want to distribute them.
It will take care of all the cases possible.
how you pretty well know how to distribute them..
OA : 9C3=84
This approach can be followed any where where you have a situation of distributing the same things to different objects. It is called as Partition method or one-zero approach.
I hope you like this approach. :)
@Vipul24 said:This is how we approach these questions:this is case where we need to distribute 6 similar things between 4 different objects(S
manochaaaa _____/\_____