Project YES
January 17, 1998

8:15am 115 VinH: Instructional Team Meeting.

8:30am 16 VinH Registration

9:00am Intro and Team Assignments -- 16 VinH

9:15am Round 1 10:00 Groups Rotate

10:05am Round 2 10:50 Groups Rotate

10:55am Round 3 11:40 All students return to VinH 16

11:45am Wrap Up & Evaluations

Counting Workshop

(A) Counting Ice-Cream

If there are three different flavors, easy to see there are 3 1-flavor coves, but how many 2-flavor cones are there?
DRAW "V", "C" and "S" for Vanilla, Chocolate, and Strawberry on board

Four flavors, V, C, S, and P for Pistachio. How many 2-flavor cones? Are you sure we have counted them all? How many 3-flavor cones?

Now suppose there are 5 flavors, C, V, S, P, and . In groups, work on: how many 2-flavor cones? 3-flavor cones? 4-flavor cones? (If you finish early, work on 6 flavors ("A" thru "F" or "1" thru "6").

Show on board: If there are five flavors, and we have a two flavor cone, we have 5 choices for the first cone, and FOR EACH of those choices, we have 4 other choices. This makes 20. But wait! "CV" and "VC" are both a vanilla-and-chocolate cone, yet we counted it twice -- in fact it's easy to see we counted everything exactly --> -- twice. So 20/2 and hence 10.

(B)Coin Game.

Draw on board, a stack of coins, starting with 12 coins. Each player gets to take 1 or 2 or 3 or 4. Whoever takes the last one, wins.

Have students try this with paper clips. Explain how instead of stacking (which is hard with pennies to go as high as 12) we'll put them in a row (one can decide on "top" and "bottom" of the pile, but that's not important (why?)).

Now pose the question: "If you get to go first, it is actually possible for you to choose the right number of coins to take away, which guarantees that you will win. How many coins should you remove if you get to go first? Can you prove this?

Have students work in groups on this.

Show solution on board (or have them present iff there is a mature, strong group with strong communication/presentation skills.

Let's work backwards. Note if we leave our "opponent" 5 coins, then we win. If they take 1, we take the last 4; if they take 2, we take the last 3, and so on. The ending of the same look like:

THIRD TO LAST STEP (n-2): I leave only 5
SECOND TO LAST STEP (n-1): He/she takes 1,2,3, or 4, leaving 4,3,2, or 1.
LAST STEP (n): I take all the remaining ones, and win

What about the 4th-to-last step? She/he must have left 6 or 7 or 8 or 9 (at the end of their turn0 for me to be able to leave only 5 at the end of my following turn.

If I want to make sure they leave 6,7,8, or 9 in the 4th to last step, then I want to leave exactly 10 in the 5th to last step. But if I go first, then the 5th-to-last step can be my first step: in my first step I take away 2, leaving 10, and then I will win!

Extra Time? (Very unlikely!)

License Plates.

How many are there of type ABC123? Enough for CA? Enough for the U.S.? Answer: 26^3 times 10^3 or 17,576,000. This is certainly not enough for all of the U.S.! It used to be enough for CA but the new plates are of the form 1ABC123, hence have ten times as many new license plates (notice the total (unless old plates are made illegal) is now 11 times as many!)

Adding up 1+2+3+4 = 10. What is 1+2+3+4+...+100? Gauss' remarkable trick.

[picture on board here of
1   +  2 +  3 + ... 98 + 99 + 100
100 + 99 + 98 + ...  3 +  2 +   1
---------------------------------
101 +101 + 101+... 101 +101 + 101

=100*101 but we counted twice so (1/2)*(100)*(101), or (1/2)(10,100)
or 5,050

Also, this is the same as "101 choose 2" which we know is 101 times 100 over 2, same result as Gauss'!