6 votes 6 votes How many solutions are there for the equation $x+y+z+u=29$ subject to the constraints that $x \geq 1, \: \: y \geq 2 \: \: z \geq 3 \: \: and \: \: u \geq 0$? 4960 2600 23751 8855 Discrete Mathematics ugcnetcse-dec2015-paper2 discrete-mathematics combinatory + – go_editor asked Aug 8, 2016 • recategorized Nov 8, 2021 by soujanyareddy13 go_editor 3.5k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 6 votes 6 votes x+y+z+u = 29 this is combination with repetition... x>=1,y>=2,z>=3,u>=0 so allocate this to directly.. 29-(1+2+3+0) = 23 (23+4-1)C23 = 2600 This problem is similar to distributing 29 similar ball into 4 numbered boxes.. papesh answered Aug 8, 2016 • selected Oct 7, 2016 by papesh papesh comment Share Follow See all 2 Comments See all 2 2 Comments reply Satya narayana commented Jan 26, 2019 reply Follow Share could you explain why are 29 distinct numbers are considered as 29 similar ball? 1 votes 1 votes pass_i0n commented Dec 16, 2019 reply Follow Share can u explain this not getting it at all :( 0 votes 0 votes Please log in or register to add a comment.
4 votes 4 votes 29 - 6= 23 (23 + 3) C3 = 26C3 = 2600 B is answer Prashant. answered Aug 8, 2016 Prashant. comment Share Follow See 1 comment See all 1 1 comment reply pass_i0n commented Dec 21, 2019 reply Follow Share why you have added 23+3 ?? may be a stupid doubt :p 0 votes 0 votes Please log in or register to add a comment.
