Counting problems are extremely common in math contests. For example:

• MOEMS Division 1 Problem 3B (2017): The first 40 odd counting numbers are written. How many times does ‘3’ appear as a digit?

There are multiple ways to solve this problem. The first way is to list out the first 40 odd numbers and count the number of 3s.

The second way to solve this problem is to use a technique called Casework counting. Casework counting divides the problem into multiple “cases” and evaluates each one. For this problem, we will divide the problem into two cases: 3s in the tens place and 3s in the ones place.

Tens place: There will be a 3 in the tens place for the odd numbers in the 30s (31,33,35,37,39).

Ones place: Since 79 is the 40th odd counting number, there will be one 3 in the ones place for every 10 numbers, starting at 3 (3,13,23,33,43,53,63,73).

In total, there are 13 3s

• Timothy rolls two dice. How many ways are there for the sum of the two dice to be at least 3?

There are 6 possibilities for the first die and 6 possibilities for the second die. Thus, there are 36 possibilities to roll the dice.

The minimum possible sum is rolling two 1s, which results in a sum of 2. There is 1 way to do this.

If we roll anything more than a 2, the conditions of the problem are satisfied. Since there are 36 total possibilities and 1 way to not satisfy the conditions, there are 36 - 1 = 35 ways for the sum of the two dice to be at least 3

For this problem, we used a technique called Complimentary counting. Complimentary counting involves subtracting the ways that the conditions of the problem are not satisfied from the total number of possibilities.

Complimentary counting can be used when casework is tedious. For this problem, to use casework, we would need to calculate the number of ways to roll a sum equal to 3,4,5,6,7, etc. It is much easier to use complimentary counting in this case.

Complimentary counting can be used in addition to casework counting, too. For example, if the problem asked for the number of ways to roll at least a 4, we could split the problem into two cases: rolling a 2 and rolling a 3. We would subtract the sum of those cases from 36 to get the answer.