Question

$A5$ digit number is formed by using the digits $0,1,2,3,4$ and $5$ without repetition. The probability that the number is divisible by $6$ is

• A. $8\%$
• B. $17\%$
• C. $18\%$
• D. $36\%$

Answer 1

The correct option is C $18\%$

According to the question,

Total $5$ digit numbers formed by the given digits$=5\ast 5!$

For divisibility by $6,$ the numbers must be divisible by both $2$ and $3$.

Only numbers formed by digits

$\left(i\right)$ Case 1: (0,1,2,4,5) For divisibility by $2$,......

If $0$ is placed at the end, the numbers of ways$=4!=24$

If $0$ is not place din the end, then their are two ways to select the last digit$(2,4).

Number of ways$=2C1\ast 3C1\ast 3!$ $=36$

For this case total number of ways$=24+36=60$

$\left(ii\right)$ Case 2: $(1,2,3,4,5)$ Again, the last number can be chosen in $2C1$ways and the remaining $4$ numbers can permute in $4!$ ways.

The required number of ways for this case$=2\ast 4!=48$

Adding the number of ways in the two cases, $5$ digit numbers that are divisible by $6=108$$

Required Probability$\frac{108}{600}=0.18$

$=18$%