Question

A five-digit number divisible by $6$ is to be formed by using $0,1,2,3,4,5$ without repetition. The number of ways in which this can be done is

• A. $60$
• B. $48$
• C. $108$
• D. $216$

Answer 1

The correct option is C $108$

We need to use the divisibility rules for both $2$ (last digit is even) and $3$ (sum of digits divisible by $3$).

First we select the $5$ digits whose sum is divisible by $3$:
Sum of all given digits $=15$ which is a multiple of $3$.
We need to leave out a single digit such that the sum remains a multiple of $3$.
Thus, we have the following two cases:
Case $1$:
$1,2,3,4,5$ are chosen.
For the number to be divisible by $2$, last digit is $2$ or $4$, i.e. last digit can be selected in $2$ ways and the other digits can arrange in $4!$. Thus, the number of ways $=4!\times 2=48$.

Case $2$:
$0,1,2,4,5$ are chosen.
a) if last digit is $0$: number of ways $=4!=24$

b) if last digit is $2$: the other digits may be filled in (from left to right): $3\times 3\times 2\times 1=18$ ways.

c) if last digit is $4$: the other digits may be filled in (from left to right): $3\times 3\times 2\times 1=18$ ways.
So, the answer $=48+24+18+18=108$ ways.

Hence, (C) is correct.