AR15J wrote:
niks18 wrote:
AR15J wrote:
It is not mentioned in the ques that one cell can have only one ball. Both the solution mentioned above are based on this assumption. How did we assume it? Please help.. This assumptions are killing me
hi
AR15Jit is clear from the image that there are 3 rows. It is mentioned in the question that each row has at least 1 ball. So you have 5 balls to play with and 3 rows how are you going to distribute them?
Thanks Niks for your reply.. I understand this.. But why can't we place more than one ball in a cell. For example , to calculate -in how many ways 2 balls can be place in a row.. why the no of ways are -- 3c1*3c2? (selecting a row and then selecting two columns from that row to place two balls)
Why are we not considering the case when both the ball are in the same cell of a row? If we consider this, total cases will be 3c1*3c2+3c1
what is more, it is stated that EACH ROW NOT EACH CELL must contain at least 1 ball
so we can first place 1 ball in each row, as there are 3 rows, we are left with (5 - 3) = 2 balls
now, as the rows and balls are identical, we can place the 2 balls in 3 rows as under
4!
____
2!2!
= 6
????