Permutations and Combinations (p-2) — Maths (commerce) STD 11 Commerce / Arts — Question

For a number to be divisible by 3.
The sum of digits must be divisible by 3.
Given 6 digits are 0, 1,2, 3, 4, 5.
Sum of 1, 2, 3, 4, 5 = 15, which is divisible by 3.
∴ There are two cases of 5 digit numbers formed from 0, 1, 2, 3, 4, 5 and divisible by 3.
Either 3 is selected in 5 digits (and 0 not selected) or 3 is not selected in 5 digits (and 0 is selected)
Case I:
3 is not selected (and 0 is selected) i.e., the digits are 0, 1, 2, 4, 5.
10000’s place digit can be selected in 4 ways (as 0 cannot appear).
As digits are not repeated, 1000’s place digit can be selected in 4 ways.
100’s place digit can be selected in 3 ways.
10’s place digit can be selected in 2 ways.
The unit’s place digit can be selected in 1 way.
∴ Using multiplication theorem,
Number of 5-digit number formed from 0, 1, 2, 4, 5 (with no repetition of digits) = 4 × 4 × 3 × 2 × 1 = 96

Case II:
3 is selected (and 0 is not selected) i.e., 1, 2, 3, 4, 5
10000’s place digit can be selected in 5 ways.
1000’s place digit can be selected in 4 ways.
100’s place digit can be selected in 3 ways.
10’s place digit can be selected in 2 ways.
The unit’s place digit can be selected in 1 way.
Using multiplication theorem,
Number of 5-digit numbers formed from 1, 2, 3, 4, 5 = 5 × 4 × 3 × 2 × 1 = 120
Both the cases are mutually exclusive and exhaustive.
∴ Required number = 96 + 120 = 216