Question

A five digit number divisible by $3$ is to be formed using the number $0,1,2,3,4$ and $5$ without repetition. The total number of ways this can be done is?

• A. $216$
• B. $240$
• C. $600$
• D. $3125$

Answer 1

The correct option is A $216$

5 digit number

$0,1,2,3,4,5$

for the number to be divisible by $5$ , sum

should be divisible by $3$.

So, $sum=3k$

(i) $k=1sum=3$

no, 5 digits number can be formed with sum$=3$ without repetition

(ii) $k=2sum=6$

no. $5$ digit number without repetition can be formed.

(iii) $k=3sum=9$

no of ways=$0$

(iv) $k=4sum=12$ $\Rightarrow (5,4,2,1,0)$

no. of possible ways $=4\text{(Selecting a non zero digit)}\times 4!\text{(arrange other 4 digits})$ (since the first digit cannot be zero)

(V) $k=5sum=15$ $\Rightarrow [5,4,3,2,1)$

no. of possible ways\, $=5!=120$ (arranging 5 digits)

$\therefore$ Total no of 5 digits number divisible by ${}^{\prime }{3}^{\prime }$ that can be formed with $0,1,2,3,4,5$ without repetition are $120+96=216$