UPSC NDA Math Model Paper 3

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