How many 3-digit odd numbers can be formed by using the digits 1, 2, 3, 4, 5, 6 when the repetition of digits is allowed ?

Concept:

Fundamental Principle of Multiplication:

Let us suppose there are two tasks A and B such that the task A can be done in m different ways following which the second task B can be done in n different ways. Then the number of ways to complete the task A and B in succession respectively is given by: m × n ways.

Fundamental Principle of Addition:

Let us suppose there are two tasks A and B such that the task A can be done in m different ways and task B can be completed in n ways. Then the number of ways to complete either of the two tasks is given by: (m + n) ways.

Calculation:

Here, we have to find how many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 such that repetition of digits is allowed

In order to form a 3-digit odd number using the digits 1, 2, 3, 4, 5, 6 the unit's digit can be filled with 1, 3 or 5 only.

So, number of ways to fill the unit's digit = 3

The number of ways to ten's digit = 6

Similarly, the number of ways to fill the hundred's digit = 6

So, the total number of required numbers = 6 × 6 × 3 = 108