Wednesday, January 21, 2015

Pascals Triangle

(Click on the comic if you can't see the full image.)
(C)Copyright 2015, C. Burke.

The 14C3 gifts of Christmas?

This occurred to me a while ago, but after Christmas, and I decided that it couldn't wait until December. (And I'd likely forget about it.)

While coming up with mathematical formulas and computer code for calculating this number, I overlooked a a very reliable reference tool: Pascal's Triangle.

It has more uses than simply expanding polynomials because of its many properties.

Some of these properties are as follows:

  • The "zeroth" element of each row is the number 1.
  • The first element of each row is the number of the row (keeping in mind that the top row is Row 0).
  • The second element of each row is are the consecutive triangle numbers, the sum of the consecutive numbers before it.
  • Which makes the third element of each row the sum of the consecutive triangle numbers.
  • And, finally, the position of each element in Pascal's Triangle corresponds to the number of Combinations designated by the notation nCr where n is the row and r is the element of that row.

    What this means is that for any given day in that song (The Twelve Days of Christmas):

  • nC1 refers to the day we're up to. On Day 7, 7C1 is 7.
  • n+1C2 refers to the total number of gifts given on the nth day. On Day 7, 8C2 is 28.
  • n+2C3 refers to the total number of gifts given altogether up to the nth day. On Day 7, 9C3 is 84.

    Applying this to the 12th day of the song:

  • 12C1 is 12.
  • 13C2 is 78.
  • 14C3 is 364.

    And I could've been finished a whole lot sooner. But I wouldn't have gotten a recursive comic out of that.

    One last thing: Did you ever try to make a poster of Pascal's Triangle? Have your students tried to do it for a math fair? The numbers start to get really big in the middle, really fast. (That's a post for another day.) But that's also why the poster in this comic is so large! Otherwise, it wouldn't be readable (and I had to modify the "364" so you could find it easily!).

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