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(B) The set of digits is $\{0, 1, 2, 3, 4, 5, 6\}$.The total number of $7$-digit numbers that can be formed is $7! - 6! = 6 	imes 6! = 4320$.$A$ numb

Vedclass · Accepted Answer

$\frac{13}{45}$

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(B) The set of digits is $\{0, 1, 2, 3, 4, 5, 6\}$.The total number of $7$-digit numbers that can be formed is $7! - 6! = 6 	imes 6! = 4320$.$A$ number is divisible by $4$ if its last two digits form a number divisible by $4$.Case $1$: The last two digits do not contain $0$. The possible pairs are $\{12, 16, 24, 32, 36, 52, 56, 64\}$. There are $8$ such pairs.For each pair,the remaining $5$ digits can be arranged in $5! - 4!$ ways (excluding $0$ at the first position).So,$8 	imes (5! - 4!) = 8 	imes (120 - 24) = 8 	imes 96 = 768$.Case $2$: The last two digits contain $0$. The possible pairs are $\{04, 20, 40, 60\}$. There are $4$ such pairs.For each pair,the remaining $5$ digits can be arranged in $5!$ ways.So,$4 	imes 5! = 4 	imes 120 = 480$.Total favorable outcomes $= 768 + 480 = 1248$.The probability $P = \frac{1248}{4320} = \frac{13}{45}$.

Consider the set of all $7$-digit numbers formed by the digits $0, 1, 2, 3, 4, 5, 6$,each chosen exactly once. If a number is randomly drawn from this set,the probability that it is divisible by $4$ is

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