We spent some time on a few more problems today. You already had the skills to solve them, but I wanted to present a different perspective.

One problem was the following:

# Example 1

Mr. Grasley has 5 different varieties of tea at home. He can bring any of them to work with him. How many different combinations of tea could he bring?

# Solution 1

Using combinations, we can count separate cases depending on the number of varieties he brings.

5 varieties:

4 varieties:

3 varieties:

2 varieties:

1 varieties:

0 varieties:

Summing these we get 32 different combinations, including the case in which he brings no tea to work.

# Solution 2

Instead of counting each case separately, we can instead realize that Mr. Grasley has to make 5 choices: for each variety of tea, he must decide whether to bring it to work or leave it at home. Using the Fundamental Counting Principle, there are

different combinations of choices.

# In general

When we are counting all of the combinations of distinct items, we have

# Example 2

We also looked at counting divisors. Every whole number is made up of a unique prime factorization, and all of its divisors are the products of combinations of those factors.

For example, let’s consider the number 1960. Rewriting it as a product of primes we have

To create a divisor, we must choose how many “copies” of each prime factor to include. For 2, there are four possibilities: 0, 1, 2, and 3 “copies”, corresponding to the numbers , , , and . Similarly there are two possibilities for 5, and three possibilities for 7. This yields different divisors for 1960. This includes the case .

# Homework

Page 287 #6, 11, 12, 13.