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We have $10$ numbered balls and $3$ boxes with capacities: $5$, $3$ and $2$ balls. With how many ways can we put the balls in the boxes? The boxes are distinguished.

I thought that it is like that:

$\binom{10}{5} \cdot \binom{5}{3} \cdot \binom{2}{2}$.Could you tell me if it is right?

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Your solution is correct. –  Suhas M Apr 26 at 12:14
    
Ok,thank you!!! –  evinda Apr 26 at 12:41
    
@evinde Just to get clear with the given problem. Is there also a distinction between the (numbered) balls ? –  calculus Apr 26 at 12:44
    
The boxes are distinguished..So,is the formula wrong?? –  evinda Apr 26 at 12:48

1 Answer 1

up vote 3 down vote accepted

Yes, indeed, your answer is correct.

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Nice,thank you amWhy !!!! –  evinda Apr 26 at 12:42

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