number of solutions

Question

Number of non negative integer solutions such that $x + y + z = 17$ where $x>1,\ y>2,\ z>3$

Comments

$45$?

---

45 seems to be correct answer

Answer 1

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 ..........x_n  =  r   where each of x_i  >= 0  is given by  ^n-1+rC_r  

So here  n = 2 , r = 8.

So required number of solutions  =   ^3-1+8C₈     =  ¹⁰C₈

                                                                     =  ¹⁰C₂

                                                                     =  45

Comments

@habib

what if it was x < 6 , y<5 , z <4

Answer 2

$x+y+z=17,x>1,y>2,z>3$

$x+y+z=17,x>=2,y>=3,z>=4$

$x^{'}=x+2,y^{'}=y+3,z^{'}=z+4$

$x^{'}+y^{'}+z^{'}=17-2-3-4=8$

Number of solution=$\binom{8+2}{2}=45$

Answer 3

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