Given ,
x + y + z = 17 where x > 1 , y > 2 , z > 3 (or) x >= 2 , y >= 3 , z >= 4.
So we can rewrite the given equation as :
(x + 2) + (y + 3) + (z + 4) = 17 where x,y,z >= 0
==> x + y + z = 8
Now we know :
Number of non negative integral solutions of x1 + x2 ..........xn = r where each of xi >= 0 is given by n-1+rCr
So here n = 2 , r = 8.
So required number of solutions = 3-1+8C8 = 10C8
= 10C2
= 45
x+y+z=17
where x>1 , y>2 , z>3
we know a+b+c=r , where a,b,c>=0 then solution would be $\binom{r+2}{r}$
so make it in that way let x-1=a y-2=b z-3=c
Now (x-1)+(y-2)+(z-3)=17-6=11
so here we made a+b+c=11 so it would give us $ \binom{13}{11}$
which is 78
64.3k questions
77.9k answers
244k comments
80.0k users