The addition principle

Use of the multiplication principle can be thought of as a “direct” method to count. However, this may not work for certain situations, especially when step 1 affects subsequent steps differently.

In such situations, we can consider using the addition principle, where we split our event up into disjoint cases.

Technique: the addition principle

Suppose we can break an event E up into two cases: case 1 or case 2 (with no overlap).

If there are m ways for case 1 to occur and n ways for case 2, then the total number of ways for E to occur is m+n.

For example, consider the earlier example of creating five-digit number using numbers from 1,2,3,4,5,6,7. This time however, we do so with repetitions not allowed. Parts (a), (b) and (c) can be solved using the multiplication principle (try it!).

However, to create a number that must be greater than 30,000 and odd, we now run into a problem unlike before.

If we chose an even number to start with (4 or 6), when we fill in the last digit, we could go with 1,3,5 or 7. However, if we chose an odd number to start with (3, 5 or 7), we have less options for the last digit (since one of 3,5,7 is used up of the possible choices of 1,3,5,7).

The addition principle is then ideal for such situations where we break our event up into cases, as can be seen in part (d) of the following example.

Example: forming a number without repetition

Question

How many ways are there to form a five-digit number satisfying the following conditions using only the numbers from 1,2,3,4,5,6,7. Repetitions are not allowed.

  1. No restrictions.
  2. The number must be greater than 30,000.
  3. The number must be odd.
  4. The number must be greater than 30,000 and odd.

Solution

Step 1: there are 7 ways to fill in the first digit.
Step 2: there are 6 ways to fill in the next digit.

Repeating for all 5 digits, required number of ways =7×6×5×4×3=2,520

Step 1: there are 5 ways to fill the first digit (3,4,5,6,7).
Step 2: there are 6 ways to fill in the next digit.

Required number of ways =5×6×5×4×3=1,800

Step 1: there are 4 ways to fill the last digit (1,3,5,7).
Step 2: there are 6 ways to fill in the next digit.

Required number of ways =4×6×5×4×3=1,440

Case 1: Start with an odd number (3,5 or 7).
We then have 3 ways to fill in the last digit.
Number of ways =3×3×5×4×3=540.

Case 2: Start with an even number (4 or 6).
We then have 4 ways to fill in the last digit.
Number of ways =2×4×5×4×3=480.

Required number of ways =540+480=1,020