Question

A 5 digits number divisible by 3 to be formed by the numerals 0, 1, 2, 3, 4 & 5 without repetition. The total number of ways this can be done is:
A. 3125
B. 600
C. 240
D. 216

Answer 1

Hint: This question is solved with the help of multiplication principle, which states that if an event can occur in m different ways, following another event which can occur in n different ways, then the total number of occurrence of the events in the given order is $m \times n$.
The above principle can be generalised for any finite number of events.

Complete step by step solution: A five digit number is formed by using digits 0, 1, 2, 3, 4 and 5 which is divisible by 3.
A number is divisible by 3 only when the sum of digits is a multiple of 3.
There are two ways to form a 5 digit number using the digits 0, 1, 2, 3, 4 and 5 which is divisible by 3
Case 1: Using the digits 0, 1, 2, 4, 5
In this case, the first place cannot be filled with 0 as it will become a 4 digit number therefore there are only 4 ways to fill the first place. On the second place also there are 4 ways to fill it. As repetition is not allowed, we cannot fill the same number again.
Therefore, the number of ways = $4 \times 4 \times 3 \times 2 \times 1 = 96$
Case 2: Using the digits 1, 2, 3, 4, 5
In this case, there are 5 ways to fill the first place, 4 ways to fill the second place and so on.
Therefore, the number of ways = $5 \times 4 \times 3 \times 2 \times 1 = 120$
Now, the total number of ways is the sum of both cases.
Hence, total number formed = 120 + 96 = 216
∴ Option (D) is correct.

Note: A permutation is an act of arranging the objects or numbers in order. Combinations are the way of selecting the objects or numbers from a group of objects or collection, in such a way that the order of the objects does not matter.
The formula for permutation is ${}^n{P_r} = \dfrac{{\left| \!{\underline {\,
  n \,}} \right. }}{{\left| \!{\underline {\,
  {n - r} \,}} \right. }}$