Find number of ordered triplets of positive integers which are solutions to the equation x+y+z=100
ans:4851
1This can be easily solved using binomial permutations.
Each number can take values between 1 to 98 with the sum being 100
This is the coefficient of x100 in (x+x2+x3+.........x98)3
I am not good at Binomial Th., So plz giv this coefficient, someone!!!
11Basically its the number of ways in which 100 objects can be divided in 3 boxes - each box getting at least 1 object!
62you want ordered triplets..
so x>=y>=z
now take y=x+a, z=y+b
so z= x+a+b
x+y+z=100
=> 3x+2a+b=100 where, a, b>=0
also, x>=1
Now finish the proof from here...
1@Nishant byah.....
i don't understand....." x>=y>=z" this condition...plz explain
62Sorry my wrong.. i was solving for the number of distinct solutions... :P (That is why i put x>=y>=z)
I din read the Q correctly...
For this question,
x=a+1
y=b+1
z=c+1
this we do so that a, b, c>=0
So we have to find a+b+c=97
which is same as partitioning 97 by 2 lines
which is 99C2 = 99x49