# The Principles of Pascal’s Triangle, Part II

Jun 26, 2001 2:32 AM

This week we’ll apply the principles of Pascal’s Triangle to tickets with three different sized groups on them. For an example, let’s take the 16 spot ticket with two groups of 3, three groups of 2, and four kings. If you remember from last week, we use the second row of the triangle to break out the three spots: A one way 6, a two way 3, and a one way “nothing.” We use the third row (because there are three groups of two) to break out the deuce ways: A one way 6, a three way 4, a three way 2, and a one way “nothing.” For the same reason, we use the fourth row to break out the one spot ways: A one way 4, a four way 3, a six way 2, a four way one spot, and a one way “nothing. “In our Keno shorthand we write these breakouts thusly:

 1/6Â Â 2/3Â Â 1/0 1/6Â Â Â¾Â Â 3/2Â Â 1/0 Â¼Â Â 4/Â Â 6/2Â Â 4/1Â Â 1/0

Proceeding as we did last week, we take the first two breakouts and arrange them on a grid, multiplying the numerators and adding the denominators:

 Â Â 1/6Â Â 2/3Â Â 1/0 Â Â *****Â Â ******Â Â ******* 1/6Â Â *1/12Â Â 2/9Â Â 1/6 3/4Â Â *3/10Â Â 6/7Â Â 3/4 3/2Â Â *3/8Â Â 6/5Â Â 3/2 1/0Â Â *1/6Â Â 2/3Â Â 1/0

The next step is to construct a second grid, (above) using the results of the first calculation for the top row, and the one spot breakout for the vertical column.

Including the 16 spot, but ignoring the now meaningless 1/0, we find the expected 511 total ways, once they are summed up. Easy, once you know the technique!

Well, that’s it for now. Good Luck! I’ll see you in line!