If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls, is

$55 over 33 open parentheses 2 over 3 close parentheses to the power of 11$
$55 space open parentheses 2 over 3 close parentheses to the power of 10$
$220 space open parentheses 1 third close parentheses to the power of 12$
$22 space open parentheses 1 third close parentheses to the power of 11$

Solution

55 over 33 open parentheses 2 over 3 close parentheses to the power of 11

There seems to be ambiguity in this question. It should be mentioned that boxes are different and one particular box has 3 balls.
Then, number of ways = $fraction numerator straight C presuperscript 12 subscript 3 space straight x space 2 to the power of 9 over denominator 3 to the power of 12 end fraction space equals space 55 over 3 space open parentheses 2 over 3 close parentheses to the power of 11$

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