A new slant on
Pascal’s (not
again!) Triangle

Feb 15, 2005 9:31 AM

Last week we looked at Pascal’s Triangle. And because most readers were probably thoroughly confused, we’ll revisit the thing this week and, hopefully, find more ways to win. For reference here it is again:

Let’s consider the simple way ticket comprised of two groups of three numbers and two groups of two numbers, for a total of 10 numbers. We know, from last week’s discussion, that since this ticket has four groups on it, if we sum the numbers on the fourth row of Pascal’s Triangle and subtract 1, we’ll have the total number of ways on the ticket.

Indeed, there are 15 total ways on the ticket, to wit: a one-way 10, a two-way 8, a two-way 7, a one-way 6, a four-way 5, a one-way four, a two-way 3, and a two-way 2.

Since this ticket has mixed groups on it, we cannot directly ascertain the number of each way from Pascal’s Triangle. However, using the bridge — an old system for checking keno tickets, — we can rapidly determine the groups on the ticket.


        1/6     2/3






To use the bridge system we must split the ticket into two slices: In this case we’ll use twos and threes. On the top row you’ll see that we are breaking out the threes, since there are two groups of three there is obviously a one-way six and a two-way three. On the first column, we are breaking out the deuces, since there are two deuces on the ticket. There is one four and a two-way two on this ticket (reading down the first column.) Although these look like fractions, they are not fractions in the traditional sense.

To use the bridge we have to multiply the numerator and add the denominator. For instance, in the column 1/6 and the row 2/2 we will have the result 2/8, and we’ll find upon sober reflection that there is a two-way eight on this ticket!

When we fill out the table, we’ll find that all fifteen ways are accounted for:


            1/6         2/3


1/4     1/10         2/7


2/2     2/8         4/5


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