Lesson plan entry: “Hand out worksheet packets and have students staple before starting. They know what to do.”
Sounds simple enough! Four numbered sheets, eight total pages, printed front and back. What could go wrong?
Do you know how many possible combinations four pieces of paper can be arranged for stapling? I believe it’s something like 1024 possible variations (not counting upside down). Fortunately, I had only 32 to check and fix.
Math wizards welcome to provide the number of actual variations possible – upside down or not.
4 comments:
Let's see. It has been a while since I have needed combinatorics.
Each of the four sheets can be put in place with the front or the back showing (2). Either of those orientations can then be put upside down or right side up (2x2=4). Multiply that by the four sheets and you get sixteen, so in essence, this is asking how many different ways can you can make a four-page packet if you have sixteen sheets to choose from, no repeats.
That would be 16x15x14x13, if I am correct, which totals 43,680 different packets. Lucky you only had 32 students!
I don't think that quite does it. You don't really have 16 independent sheets, you have 4 sheets with lots of options for each sheet. I think it would be like this:
First, the four different sheets can be arranged top-to-bottom in 4x3x2x1=24 different ways.
Then, each sheet can have either side on top: 2x2x2x2=16 combinations.
Then, each sheet can be upside down or right side up: 2x2x2x2=16 again.
Then, each sheet could be turned to portrait or landscape orientation, but it's hard to believe the students would try to mix those, so I'll leave that out.
Finally, there are a nearly-infinite number of places to put the staple. But we will assume the students have enough experience to know that staples belong in the corner, so there are 4 corners to choose from.
The calculation so far:
24x16x16x4=24,576 possible packets.
BUT some of these packets are the same, because real-life packets can be rotated. For instance, a right-side-up packet stapled at the bottom right would be identical with an upside-down packet stapled at the top left, wouldn't it? And then, if you don't distinguish between the sides of the staple, you will have flip-symmetry as well.
24,576÷2÷2=6,144 distinguishable possibilities?
--Denise
Actually, all of your calculations are a little suspect. You have 32 students. Each student has 4 papers, each printed front and back. With all the possible combinations taken into account I come up with...
ONE! As in One Big Mess! Leave it to the students to make it.
four choices for the first sheet, times two faces, times two up and down orientations, times
three choices for the second sheet, times two faces, times two up and down orientations, times
two choices for the third sheet, times two times two, times
one choice for the last sheet, times two times two:
(4*2*2)(3*2*2)(2*2*2)(1*2*2)= 4!*2^8 = 6144
What comes next is interesting. We divide by 4 (each packet exists upside down and right side up and flipped and not flipped), but then multiply by 4 (like Denise said, staple can go in 4 places. You could argue that only two of those spots are reasonable, so 3072 would be an equally ok answer)
So I agree, almost, with Denise, and generalize to n sheets of paper: (n!*4^n)/2
Jonathan
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