# Pascal's Triangle (all digits, last digit, and combinametrics)

Pascal's Triangle is created by starting with only 1s in each of the two perimeter diagonals of a triangular array. Fill in this array by adding the two numbers above each cell. The numbers start out small such as 1 2 1 on the second row and 1 3 3 1 on the third row and 1 4 6 4 1 on the fourth row but then they explode. There are endless patterns to be found and that is what we will explore next week.

Write lightly because when you make a mistake, every cell in the triangle below that cell will need to be erased. The cells are identified by row and element. The seed number “1” on the left is the zeroth element so the first element is the row number. For example, in the row 121, this is row 2; in the row 1 3 3 1, this is row 3; in row 4: 1 4 6 4 1, the 2nd element is 6. See page 1 of the pdf for further detail.

The first few rows you can add mentally. However, once you get into the double, triple digits and beyond, add your numbers vertically.

With the K-2nd graders we did a challenge where they recorded only the units digit for each cell. So, 10 record a 0, 11, record a 1, 12 record a 2, 13 record a 3 and so on. This is challenging since one mistake will cause each cell in the triangle below that cell to be incorrect.

I gave answers in a lower row so check to make sure you are correct. If not, find your mistake, erase, and re-calculate.

START WITH A SEED NUMBER OTHER THAN 1 ALONG EACH DIAGONAL EDGE OF PASCAL’S TRIANGLE ON THE LAST PAGE OF THE PDF AND EXPERIMENT WITH WHAT HAPPENS TO EACH CELL.

With the 5th graders and up we explored my conjecture that all prime numbered rows: 2,3,5,7,11,13 and so on have a special property. Each number except for 1, will be a multiple of that prime number. So in row 7, the numbers 7 21 35 35 21 7 are all multiples of 7; whereas in row 8, the numbers 8 28 56 70 56 28 8, only a few numbers are multiples of 8. Test this theory on rows 11, 13 and 17. Why does this work? We explored combinametrics to prove my Prime Conjecture:

The number of combinations of "n elements taken r at a time" can be found by using the notation and formula (a combination of ABC is the same as BCA so order does not matter):

NCR=(N!)/((N-R)!R!)

n! or n factorial means n(n-1)(n-2)(n-3)... (1). So, 4! = 4 x 3 x 2 x 1 = 24

Using Pascal's Triangle, if you look at the 8th row and the 6th element (remember, the first element is the row number 8, not 1), we can calculate that row by using the combinametrics formula. See the pdf for a full explanation of this formula.

Check out my Pascal’s solution to 29 rows.

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