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(C) The total number of ways to form a three-digit number using the digits $1, 2, 3,$ and $4$ (without repetition) is $P(4, 3) = 4 	imes 3 	imes 2 =

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$\frac{1}{2}$

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(C) The total number of ways to form a three-digit number using the digits $1, 2, 3,$ and $4$ (without repetition) is $P(4, 3) = 4 	imes 3 	imes 2 = 24$.$A$ number is divisible by $3$ if the sum of its digits is divisible by $3$.Possible sets of $3$ digits from ${1, 2, 3, 4}$ whose sum is divisible by $3$ are:$1) \{1, 2, 3\}$ (Sum $= 6$,which is divisible by $3$)$2) \{2, 3, 4\}$ (Sum $= 9$,which is divisible by $3$)For each set,the number of ways to arrange the $3$ digits is $3! = 6$.Total favorable outcomes $= 6 + 6 = 12$.Required probability $= \frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{12}{24} = \frac{1}{2}$.

$A$ three-digit number is formed by using the digits $1, 2, 3,$ and $4$. The probability that the number is divisible by $3$ is:

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