Question

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

• A. 8%
• B. 17%
• C. 18%
• D. 36%

Answer 1

Answer: (C)

18%

Total $5$ digit numbers formed by the given digits$=5\times 5!=600$

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

Only numbers formed by digits $\left(i\right)(0,1,2,4,5)$ and $\left(ii\right)(1,2,3,4,5)$ are divisible by $3$.

(i) CASE 1: $(0,1,2,4,5)$ For divisibility by $2$,

(a) if $0$ is placed at the end, then number of ways$=4!=24$

(b) if $0$ is not placed in the end, then there are two ways to select the last digit$(2,4)$. Number of ways$={}^2{C}_1\times {}^3{C}_1\times 3!=36$ For this case, total number of ways$=24+36=60$

(ii) CASE 2: $(1,2,3,4,5)$ Again, the last number can be chosen in ${}^2{C}_1$ ways and the remaining $4$ numbers can permute in $4!$ ways. The required number of ways for this case$=2\times 4!=48$

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

Required Probability $=108600=0.18=18\%$