Reflections from Teaching: Multiple Strategies Converge into One
“Then all the colors will bleed into one.” -U2
Yesterday was just one of those days when you realize that you are in the right profession. I had an amazing and unscripted teaching moment while leading my pre-service teachers through a guided-discovery lesson in permutations/combinations.
They had split themselves up into small groups and I gave each group 5 unique playing cards. Their task was to determine the number of ways to select 3 cards and arrange them in a row. (Note: before this task had started, we had already solved the problem of selecting 3 cards from 5 and not ordering them. This task was upping the ante and asking how the problem changes when order becomes important.) It was unscripted in the sense that I had left them to their own devices – they were certainly expert enough that this task fell within their ‘zone of proximal development’ and I was confident enough in my abilities with permutations/combinations to connect the dots.
The Most Specific
I walked around to chat with the various groups to determine how they were thinking about the problem. The first group said they thought it was 5x4x3=60 ways by explaining that for the first card, they had 5 choices, then for the second card they had 4 choices, and the last card there were 3 choices. Awesome.
The Most General
The next group was playing around with 6s. So I asked them why they were thinking about 6s. They said they knew they had to order 3 of the 5 cards, and to order 3 cards this is 3!=3x2x1=6, so they were trying to figure out how many groups of 6 they had. I thought this was really interesting and offered to them that they already knew how many groups of 6 there were (ie. they knew how to determine how many way to choose 3 cards out of 5). It wasn’t too much of a push to get them to see they had 10 groups of 6, or 60 ways to order 3 of 5 cards.
In all honesty, this one caught me off-guard – but turned out to be the most interesting conversation! I had no idea where the 6s were coming from and I was glad I asked them to share their thoughts, otherwise I may not have connected the dots to see they were actually thinking about 3!
For the final group, they had yet a different strategy. When selecting 3 of 5 cards, we can use the formula 5C3 = 5!/(2!3!), and they knew that the 5! in the numerator orders all cards; while the 2! and 3! in the denominator remove the ordering for the group of 3 cards, and the 2 left-over cards. This group had originally come up with 5!/3! ways; however, a slight chat about which set they wanted to order (did they want to order the group of three cards or the left-over 2 cards?) had them change their answer to 5!/2!=60 ways.
Converging to One Idea
Now that I had had an opportunity to discuss with each group and see they ways that they were thinking about the problem, I led a class discussion allowing each group to share their strategy for thinking about the problem. I filled in any gaps and wrote the three strategies on the board. It was now time to show how each strategy was actually the same idea!
I began with the ideas of the first and last groups. I asked the class to take 5!/2! and simplify to get a numerical value.
5!/2! = (5x4x3x2x1)/(2×1) = 5x4x3 = 60
Perhaps you can see that the first group’s answer to the problem is hidden within the third group’s solution! The next phase was to try to connect the second groups’d idea.
(#of ways to select 3 of 5 cards) x (# of ways to order 3 cards)
= 5C3 x 3!
= 5!/(2!3!) x 3!
= 60 ways
Now we can see all three strategies connected into one larger picture. From here we made note that to select AND order a subset of items, we can use nPr = nCr x r! and the connection between permutations and combinations was made.
It was a lot of fun to see that each group had a different way to solve the question. And for me it was a nice way to see how educational theories come into play – each group was trying to answer this new problem by starting with something they already knew. Group 1 likely saw this problem as a tree (for each of the first 5 choices, I have 4 choices, …). The second group knew how to order 3 cards so they likely altered the original question to ask “How many groups of 6 can I make?” The final group was comfortable knowing that 5!/(2!3!) removed the order from the group of 3 and the left-over 2 and likely asked “Is there a way I can alter this formula to solve the current problem?”