Question

The five-digit number divisible by 3 is to be formed using the numerals 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

Answer: (A)

216

Answer 2

First note that a number will be divisible by 3 if the sum of its digits is divisible by 3. Let us make first selection of 5 digits out of 6 given digits i.e. ⁶C₅ = ⁶C₁ = 6 ways as listed below (without arrangement). 01234, 01235, 01245, 01345, 02345 and 12345.

In all the above combinations only the 3^rd and 6^th sets are such that sum of their digits being 12 and 15 respectively are divisible by 3. It is only these two sets which have now to be arranged. 12345 can be arranged in 5! = 120 ways 01245 can be arranged in 5! – 4! = 120 – 24 = 96 ways

4! corresponds to the number which have 0 in the beginning. Thus the total number of desired numbers is 120 + 96 = 216.

Another Method:

We know that a number is divisible by 3 if the sum of digits is divisible by 3. We have 0, 1, 2, 3, 4, 5.

Numbers without zeros 5! = 120

Each will contain digits 1, 2, 3, 4, 5 whose sum 15 is divisible by 3.

Numbers with zeros. They will have one zero and 4 other digits like 1234, 2345, 3451, 4512, 5123. Out of these only one set having 4512 whose sum is 12 will be divisible by 3. Thus we have to form 5 digit numbers out of 0, 1, 2, 4, 5,

= 5! – 4! = 120 – 24 = 96

Total = 120 + 96 = 216.