 # The slant on Pascal’s Triangle

Jun 18, 2001 1:43 AM

Pascal’s triangle may be constructed by taking a piece of paper and writing down the two numbers 1 and 1 side by side. This is the first line of the triangle:

1 1

Subsequent lines are added by starting each line with a 1, and then by adding together the vertices of the triangle diagonally above, and ending each line with a 1 as illustrated.

We’ve discussed before the application of Pascal’s Triangle to elementary Keno calculations, those tickets that have all the same size groups on them. For instance, given a way ticket with 10 spots on it, grouped in five groups of two, we use the fifth row of the triangle, and reading across, we get a one way ten, five way eight, ten way six, ten way four, five way two, and a one way “nothing.” Now the “nothing” means nothing in itself, but it will serve as a place holder in further calculations.

What if the ticket is a mixed group ticket? For instance, suppose it has ten numbers, grouped 3-3-2-2? For this we need to do a two-step calculation. First, using the second line for the two groups of three, we find that we have a one way six, a two way three, and a one way “nothing.” Putting this in a fractional notation, we write:

1/6 2/3 1/0

Of course these are not true fractions in the mathematical sense, just a Keno shorthand that we use. We read 2/3 as “two way three.” Doing the same with the two groups of two, we get:

1/4 2/2 1/0

We next create a grid, with the three spot ways across the top, and the two spot ways in the leftmost column. We can then calculate the ways on the ticket by multiplying the numerator and adding the denominator to fill the grid, like this:

 Â Â Â Â  1/6 2/3Â 1/0Â Â Â ******** **** **** 1/4Â Â *1/10Â Â Â 2/7Â Â 1/4 2/2Â Â * 2/8 4/5Â Â Â 2/2 1/0Â Â *1/6Â Â Â 2/3Â Â Â 1/0

You will now find all 15 ways on the ticket displayed inside the grid lines, plus the “place holder” (1/0), which we can ignore. Next week we’ll cover tickets with three different sized groups using this method.