Question

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

• A. 216
• B. 240
• C. 600
• D. 3125

Answer 1

The correct option is A

216

We know that a five digit number is divisible by 3, if and only if sum of its digits (= 15) is divisible by 3, therefore we should not use 0 or 3 while forming the five digit numbers.

Now,

(i) In case we do not use 0 the five digit number can be formed (from the digit 1, 2, 3, 4, 5) in ${}^5{P}_5$ ways.

(ii) In case we do not use 3, the five digit number can be formed (from the digit 0, 1, 2, 4, 5) in ${}^5{P}_5-{}^4{P}_4$ = 5! - 4!= 120 - 24 = 96 ways.
(${}^4{P}_4=$ cases when $0$ is at the first position)

$\therefore$The total number of such 5 digit number = ${}^5{P}_5+({}^5{P}_5-{}^4{P}_4)=120+96=216$