Lesson 1: Combinations and Permutations

Step 1 | Step 2 | Step 3 | Curriculum Connections

Step 2: Combinations verses Permutations

Explain: Permutations and combinations are counting tools used in probability, They are used to determine the number of successful outcomes and the number of total outcomes in a given scenario.

Divide students into small groups. Explain that questions 1 and 2 above are examples of combinations. Ask students the difference between a combination and a permutation.

A combination of objects is a selection of objects where the order of these objects does not matter.

A permutation of objects is an arrangement of objects in a particular order.

Have students work on the following problems as a group, recording their work.

Question 1a: Lets imagine a combination lock comprised of three digits, the possibilities for each ranging from 0 - 9. Before we calculate how many different possible number combinations on the lock, determine if this is an example of combinations or permutations.

Answer: Contrary to its name, when you are finding possible combinations on a lock, you are actually finding the number of possible permutations. That is because the numbers are selected in a specific order.

The combination lock is an example of a permutation where repetition is allowed, meaning the occurrence of any of the events (the numbers) can be repeated (ex: the combination could be 333).

Question 1b: Ask students how they might approach finding the number of possible combinations on the lock. Would using a grid or tree diagram be of use for this kind of problem?

Answer: The number of possible outcomes is large so these strategies are impractical.

Representing the data in a formula is the most efficient way to analyze the possibilities:

The formula for a permutation in which repetition is allowed is simply n^r (n to the power of r) where n is the number of things to choose from, and you choose r of them.

Question 1c: Applying this formula, find all the possible permutations of the lock.

Answer: n^r = 10 x 10 x 10 = 1000 different permutations for a 3 digit combination lock.

Question 2: How many permutations for a 5 digit pin number, with numbers ranging from 0 - 9?

Answer: 10 x 10 x 10 x 10 x 10 = 100 000

Give each group a set of six coloured squares: one of each orange, red, blue, green, white and yellow. Ask students to pull out an orange, blue, yellow and green one (4 squares), and arrange them in a row.

Question 3a: How many ways can these four colours O, B, G and Y be ordered in a row? Create a list of the possibilities using the pieces to help you visualize.

Answer: 24

It is faster and simpler to apply a formula involving the multiplication rule to find the number of arrangements.

To calculate the number of permutations without repetition is, we use the factorial function: n! = n x (n – 1) x (n – 2)...
n! means n factorial.
Note: 0! = 1

So for our four coloured squares, the formula would be: 4! = 4 x 3 x 2 x 1 = 24.

Question 3b: Why would we multiply the four numbers in descending order?

Answer: The number of possibilities decreases after every choice, i.e.: for your first selection you have four options, for the second you have three, and so on.

Question 4a: Pull out one square of each colour so that you have six coloured squares. How many arrangements of three different coloured squares in a row can you make? In this question, you are choosing permutations (because they must be ordered in a row) of only three possibilities out of the six squares you have. Manipulate the squares and then list the possibilities.

Now we are choosing r items out of n items, so we need a new formula:

The formula for finding the number of permutations when selecting r number of items from n is: nPr = n!/ (n - r)!

People use different notations like these:

We reduce the number by dividing by the number of things being omitted — in this case, we are permutating 3 objects from six, and omitting 3. So 6! will be divided by 3!

Question 4b: Applying this formula, how many ways can three squares be arranged with six squares to choose from?

Answer: 6P3 = 6 x 5 x 4 x 3 x 2 x 1 / 3! = 6 x 5 x 4 x 3 x 2 x 1 / 3 x 2 x 1 = 6 x 5 x 4 = 120 (the equation is reduced by canceling out repetitions 3 x 2 x 1 in the numerator and denominator). There are 120 different permutations of the three colours out of six.

Question 5: Now find the number of combinations of the same three colours out of six. Remember, the order in which they’re selected doesn’t matter.

The formula for finding the number of combinations when selecting r number of items from n is nCr = n! / (n – r)!r!

So the answer to question 5 is written: 6C3 = 6 x 5 x 4 x 3 x 2 x 1 / 3!3! = 6 x 5 x 4 x 3 x 2 x 1 / (3 x 2 x 1) (3 x 2 x 1) = 6 x 5 x 4 / 3 x 2 x 1 = 120 / 6 = 20

Question 6: Why are there more permutations then combinations?

Answer: With combinations, order does not matter, and so we reduce the possibilities further by dividing by the number of things that could be in order (because we don't care about the order). This makes a much smaller number. For example there are six ways to order yellow, red and blue but only one way to combine them all.

Here are two more examples of combinations and permutations. In their groups, students first identify if the problem is an example of a combination or a permutation before solving it.

Question 7: Choose three pizza toppings from a choice of ten.

Question 8: You have Olympic athletes from 9 countries competing for gold, silver and bronze metals.

Permutations in a circle:
Sometimes, items are arranged in a circle. In this case, we no longer have a left end and a right end. The first placed item is merely a point of reference instead of having n choices. Thus with n distinguishable objects we have (n-1)! arrangements instead of n!.

Question 9: Arrange the six coloured squares in a circle. How many arrangements are there? Encourage the students to make arrangements of their squares in a circle to more concretely understand this formula.

Answer: In a line, there would be 6! = 720 such arrangements. If considered as a circular arrangement there are but 5! = 120 arrangements.