# A 5 digit number is formed using the digits 0,1,2,3,4 and 5, probability it is divisible by 6?

A $5$ 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?

I had doubts regarding this question but while writing my attempt here, I got the solution. So, I am posting it as an answer.

• I just got it. In case 1, for divisibility by 2, the last number can be chosen in 3 ways(0,2,4) – Arishta Jun 7 '17 at 4:29
• Yes, also if $0$ is chosen for the last digit, the remaining digits can be permuted $4!$ ways. – Graham Kemp Jun 7 '17 at 4:32

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

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

Only numbers formed by digits (i) (0,1,2,4,5) and (ii) (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=$2C1*3C1*3!=36$ 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 $2C1$ ways and the remaining 4 numbers can permute in $4!$ ways. The required number of ways for this case= $2*4!= 48$

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%