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?

Answer : 18%

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.