2 votes 2 votes Number of non negative integer solutions such that $x + y + z = 17$ where $x>1,\ y>2,\ z>3$ Mathematical Logic discrete-mathematics combinatory + – A_i_$_h asked Oct 9, 2017 • edited Jun 2, 2021 by Shiva Sagar Rao A_i_$_h 2.6k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply sourav. commented Oct 9, 2017 reply Follow Share $45$? 1 votes 1 votes rahul sharma 5 commented Oct 9, 2017 reply Follow Share 45 seems to be correct answer 1 votes 1 votes Please log in or register to add a comment.
Best answer 5 votes 5 votes 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 Habibkhan answered Oct 9, 2017 • selected Oct 9, 2017 by A_i_$_h Habibkhan comment Share Follow See 1 comment See all 1 1 comment reply A_i_$_h commented Oct 11, 2017 reply Follow Share @habib what if it was x < 6 , y<5 , z <4 0 votes 0 votes Please log in or register to add a comment.
6 votes 6 votes $x+y+z=17,x>1,y>2,z>3$ $x+y+z=17,x>=2,y>=3,z>=4$ $x^{'}=x+2,y^{'}=y+3,z^{'}=z+4$ $x^{'}+y^{'}+z^{'}=17-2-3-4=8$ Number of solution=$\binom{8+2}{2}=45$ sourav. answered Oct 9, 2017 sourav. comment Share Follow See all 0 reply Please log in or register to add a comment.
–1 votes –1 votes 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 Rupendra Choudhary answered Oct 9, 2017 Rupendra Choudhary comment Share Follow See all 0 reply Please log in or register to add a comment.