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:
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:
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