__Lesson Theme__: Introduction to Probability and Counting

__Prerequisite Knowledge__: Permutations and Combinations

Here’s an activity that I introduced to four groups of tenth graders in a recent unit on probability:

- Give each team a travel lock and a different hint. No communication between teams are allowed. They have two minutes to try and “break the code” and find the combination that opens the lock.
- The reveal: show the class all the different tasks they were given. Ask:
*Did each team have a fair chance of winning? Which team do you think had the highest chance of finding the combination? What about the lowest? What strategies did each team employ to try and crack the code?*

- Have students support their answers by calculating the number of possible codes for
*each*lock. Each group needs to*commit*to an answer by writing it on the board. (This ensures the type of “buy in” necessary for students to invest in the problem). - Have groups come up one at a time to explain their calculations.
- Have students find the the probability that each team breaks the code on the first try, introducing notation along the way (e.g. “P(A) = probability of event A”)

Provide minimal guidance as the groups decide on the number of combinations for each scenario. I made an exception for students who asked clarifying questions, such as:

– Are there any repeating digits?

– How many digits repeat?

The hints are open to interpretation on purpose in order to get students thinking about the sorts of constraints they would need to consider when calculating the total number of outcomes. The discussion phase of the activity provided a rich opportunity to address student misconceptions about permutations and combinations, as well as the importance of reasoning, i.e. *Does this number make sense? Is this estimate too high or too low? How does this number compare with my initial guess (intuition)? What if there were no constraints, what would the answer be?*

__The Solutions__:

TEAM GARRY – 3 digit code, repetitions allowed. This hint is not much of a hint at all. “Repetitions allowed” could mean that there may be or may not be any repetitions in the code. So, one possible answer is simply 10 x 10 x 10 = 1000. There are ten ways of picking the first, second, and third digits.

TEAM ORVILLE – 3 digit code, no repetitions. This narrows the playing field a bit. 10 x 9 x 8 = 720. Ten ways to pick the first digit, only 9 choices left for the second, then 8 for the third.

TEAM APRIL – 4 digit code, numbers 2 and 3 appear in the code.**(a)** This would be significantly easier if the ONLY numbers in the code were 2 and 3. There are only 4 possible combinations, which are easy to figure out by hand: {2323, 3232, 2233, 3322} or

4!/ (2! x 2!) = 4.**(b)** With repetition. There are 4 ways of positioning the “2” in a four digit combination. For each way that the “2” has been positioned, there are 3 ways to position the “3.” In total, we can arrange the digits “2” and “3” in 4 x 3 or 12 ways. If repetition is allowed, the total number of combinations would be 12 x 10 x 10 = 1200.

(c) Without repetition. Similar to above, except that once we have found a placement for the “2” and “3”, there are only 8 and 7 digits left to choose from respectively. The final answer would be

2 x 8 x 7 = 672

__My notes and observations__:

- Having done a previous unit on permutations and combinations, students tended to misuse the combinations formula without fully understanding
*what it actually helped us find*(e.g. 9C3 gives the total number of ways to pick three objects out of 9, when*order does not matter)* - While this did not come up in our discussions, you may choose to ask the class, “What types of combinations would you say are safe to rule out, or eliminate?” [
*EX. People are less likely to pick combinations that have just one repeating digit like 000, or consecutive sequences such as 123]* - For lower ability groups, it is advisable to go through the solutions one at a time rather than all at once. Give groups the option of revising their original solution, agreeing with another group’s solution, or arguing another groups solution after each round of explanations
- Depending on the depth of discussion, this activity can range from just 10 mins to 40 mins