MDM4U Special Combinatorics 2016-02-29

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: \binom{5}{5}=1

4 varieties: \binom{5}{4}=5

3 varieties: \binom{5}{3}=10

2 varieties: \binom{5}{2}=10

1 varieties: \binom{5}{1}=5

0 varieties: \binom{5}{0}=1

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 n 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

1960 = 2^3 \times 5 \times 7^2

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 2^0 2^1 2^2 , and 2^3 . Similarly there are two possibilities for 5, and three possibilities for 7. This yields 4 \times 2 \times 3 = 24 different divisors for 1960. This includes the case 2^0 \times 5^0 \times 7^0=1 .


Page 287 #6, 11, 12, 13.


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