1 votes 1 votes How to solve? ________________ Number of non-negative integer solutions such that x+y+z=17 where x>1,y>2,z>3 is -------- Linear Algebra system-of-equations + – ankit_thawal asked Jan 13, 2018 ankit_thawal 548 views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments Anu007 commented Jan 13, 2018 i edited by Anu007 Jan 13, 2018 reply Follow Share x+y+z=17 where x>1,y>2,z>3 x+2+y+3+z+4= 17 x+y+z= 17- 9 x+y+z= 8 8+3-1C8 = 10C8 = 45 2 votes 2 votes MiNiPanda commented Jan 13, 2018 reply Follow Share @Anu007 there is greater than sign not greater than equal to, so x must be at least 2 , y at least 3 and z at least 4 right? 0 votes 0 votes Anu007 commented Jan 13, 2018 reply Follow Share yeah...:) 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes since x>1,y>2 and z>3 so new equation will be x+1+y+3+z+4=17 =>x+y+z=8 now no. of solution of these equations are (n+r-1)C(r-1) here n=8 r=3(no. of unknowns) so 10C2=45 Ans Navneet Kalra answered Jun 28, 2018 Navneet Kalra comment Share Follow See all 0 reply Please log in or register to add a comment.