Combinatorics Through Guided Discovery, 2004a

1.2. Basic Counting Principles 13
Problem 30. Just for practice, what is the next row of Pascal’s triangle?
⇒
Problem 31. Without writing out the rows completely, write out enough of
Pascal’s triangle to get a numerical answer for the first question in Problem
8. (h)
It is less common to see Pascal’s triangle as a right triangle, but it actually
makes your formula easier to interpret. In Pascal’s Right Triangle, the element in
row n and column k (with the convention that the first row is row zero and the
first column is column zero) is ( n ).
k
In this case your formula says each entry in a
row is the sum of the one above and the one above and to the left, except for the
leftmost and rightmost entries of a row, for which that doesn’t make sense. Since
the leftmost entry is ( n 0 ) and the rightmost entry is (n ),
n
these entries are both one
(to see why, ask yourself how many 0-element subsets and how many n-element
subsets an n-element set has), and your formula then tells how to fill in the rest of
the table.
k =0 1 2 3 4 5 6 7
n =0 1
1 1 1
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
5 1 5 10 10 5 1
6 1 6 15 20 15 6 1
7 1 7 21 35 35 21 7 1
Table 1.2.5: Pascal’s Right Triangle
Seeing this right triangle leads us to ask whether there is some natural way
to extend the right triangle to a rectangle. If we did have a rectangular table of
binomial coefficients, counting the first row as row zero (i.e., n =0) and the first
column as column zero (i.e., k =0), the entries we don’t yet have are values of ( n k )
for k > n. But how many k-element subsets does an n-element set have if k > n?
The answer, of course, is zero, so all the other entries we would fill in would be
zero, giving us the rectangular array in Figure 1.2.6. It is straightforward to check
that Pascal’s equation now works for all the entries in the rectangle that have an
entry above them and an entry above and to the left.