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(C) The total number of ways to arrange the digits $1, 2, 3, 4, 5$ is $5! = 120$.For a number to be divisible by $4$,the number formed by its last two

Vedclass · Accepted Answer

$\frac{1}{5}$

Vedclass · Answer

(C) The total number of ways to arrange the digits $1, 2, 3, 4, 5$ is $5! = 120$.For a number to be divisible by $4$,the number formed by its last two digits must be divisible by $4$.The possible two-digit combinations from the set ${1, 2, 3, 4, 5}$ that are divisible by $4$ are $12, 24, 32,$ and $52$.For each of these $4$ pairs,the remaining $3$ digits can be arranged in the first $3$ positions in $3! = 6$ ways.Therefore,the number of favorable outcomes is $4 	imes 3! = 4 	imes 6 = 24$.The probability is $\frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{24}{120} = \frac{1}{5}$.

$A$ five-digit number is formed by writing the digits $1, 2, 3, 4, 5$ in a random order without repetition. What is the probability that the number is divisible by $4$?

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