p&c

Question

six balls have to be placed in the squares such that each row has atleast  one ball . no of ways of doing this is-

Answer 1

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.

Comments

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.

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ohkkk...thank you :) :)

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if i do like 2c1*4c4*2c1+ 2c2*4c3*2c1+ 2c2*4c2*2c2 =18  then whats going on wrong here

someone verify