0 votes 0 votes six balls have to be placed in the squares such that each row has atleast one ball . no of ways of doing this is- Combinatory combinatory + – shreshtha5 asked Dec 28, 2015 • retagged Jun 27, 2017 by Arjun shreshtha5 656 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 3 votes 3 votes There are 8 cells and 6 balls to place, so we can do it in $\binom{8}{6}$ . Now we need to remove two cases where toprow is left out without balls and second where last row is left out. So then it becomes $\binom{8}{6}$ - 2 = 26. UK answered Dec 28, 2015 • selected Dec 28, 2015 by Praveen Saini UK comment Share Follow See all 8 Comments See all 8 8 Comments reply Show 5 previous comments UK commented Dec 29, 2015 reply Follow Share As far as I can make out the problem is that since where these first 3 balls are being placed is not being specified 5C3 could mean that 2 cells(top row) and 3 cells(in mid row) are the remaining places or 2 cells(last row) and 3 cells(in mid row) are the remaining places or 1 cell( top row), 3 cells ( mid row), 1 cell (last row). Because you are visualizing it so you are considering only the last option. 1 votes 1 votes shreshtha5 commented Dec 29, 2015 reply Follow Share ohkkk...thank you :) :) 0 votes 0 votes sid1221 commented Jun 6, 2017 reply Follow Share if i do like 2c1*4c4*2c1+ 2c2*4c3*2c1+ 2c2*4c2*2c2 =18 then whats going on wrong here someone verify 0 votes 0 votes Please log in or register to add a comment.