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(C) The total number of three-digit numbers that can be formed using the digits $1, 2, 3,$ and $4$ without repetition is given by $^4P_3 = 4 	imes 3

Vedclass · Accepted Answer

$\frac{1}{2}$

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(C) The total number of three-digit numbers that can be formed using the digits $1, 2, 3,$ and $4$ without repetition is given by $^4P_3 = 4 	imes 3 	imes 2 = 24$. $A$ number is divisible by $3$ if the sum of its digits is divisible by $3$. Possible sets of three digits from ${1, 2, 3, 4}$ whose sum is divisible by $3$ are: $1 + 2 + 3 = 6$ (divisible by $3$) $2 + 3 + 4 = 9$ (divisible by $3$) For the set ${1, 2, 3}$,the number of arrangements is $3! = 6$. For the set ${2, 3, 4}$,the number of arrangements is $3! = 6$. Total favorable outcomes = $6 + 6 = 12$. Required probability = $\frac{12}{24} = \frac{1}{2}$.

$A$ three-digit number is formed by using the digits $1, 2, 3,$ and $4$ without repetition. The probability that the number is divisible by $3$ is:

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