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(D) Let the number of balls be $n = 5$ and the number of boxes be $k = 3$. Since the balls are distinct and the boxes are distinct, we use the Princip

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$150$

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(D) Let the number of balls be $n = 5$ and the number of boxes be $k = 3$. Since the balls are distinct and the boxes are distinct, we use the Principle of Inclusion-Exclusion to ensure no box is empty. The total number of ways to place $5$ distinct balls into $3$ distinct boxes is $3^5 = 243$. Let $S$ be the set of all possible distributions. Let $A_i$ be the property that box $i$ is empty. We want to find the number of ways such that no box is empty, which is $|S| - |A_1 \cup A_2 \cup A_3|$. By the Principle of Inclusion-Exclusion: $|A_1 \cup A_2 \cup A_3| = \sum |A_i| - \sum |A_i \cap A_j| + |A_1 \cap A_2 \cap A_3|$. $|A_i| = 2^5 = 32$. There are $\binom{3}{1} = 3$ such cases. $|A_i \cap A_j| = 1^5 = 1$. There are $\binom{3}{2} = 3$ such cases. $|A_1 \cap A_2 \cap A_3| = 0^5 = 0$. Number of ways = $3^5 - \binom{3}{1} 2^5 + \binom{3}{2} 1^5 = 243 - 3(32) + 3(1) = 243 - 96 + 3 = 150$.

Five balls of different colours are to be placed in three boxes of different sizes. The number of ways in which we can place the balls in the boxes so that no box remains empty is

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