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|dw:1440892159292:dw|
In how many different ways can we put all 5 identical balls in a 3\(\times\)3 grid . Such that each row contains was atleast one ball.
@Nnesha please remove question mark

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no mod power today sorry ;~;
think about choosing 4 blanks squares from 9, maybe
9C4
answer is 108
There are 9 C 5 = 126 ways to place the balls without worrying about whether all rows have at least one ball or not. If we focus on the cases where one row is left out, like row 3 left out, then we have 6 C 5 = 6 ways to fill up the first two rows only. So there are 3*6 = 18 ways to fill up 2 rows only leaving out one row blank The last thing to do is subtract 126 - 18 = 108
that works
how is leaving one row empty and chossing 5 out 6 and multiplying by 3 is similar to condition at least one row has one ball
Well you only have 2 options: you either have each row have at least 1 ball OR one row gets left out entirely when I calculated the "18", that's the number of ways to leave out 1 row I subtracted from 126 because of this property (number of ways where each row gets at least 1 ball) + (number of ways to leave out one row entirely) = total number of ways to fill up the slots

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