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\title{A Formula for Pascal's Triangle}

\author{Math 166: Honors Calculus II}
\date{Spring 2000}

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The sum of the numbers on a diagonal of Pascal's triangle equals
the number below the last summand. For example, $ 1 + 2 = 3$, $1 + 2 + 3 =
6$, $1 + 3 =4$, $1 + 3 + 6 = 10$, etc.
\[\begin{array}{ccccccccccc}
  &    &    &    &    &  1 &    &    &    &    &   \cr
  &    &    &    &  1 &    &  1 &    &    &    &   \cr
  &    &    &  1 &    &  2 &    &  1 &    &    &   \cr
  &    &  1 &    &  3 &    &  3 &    &  1 &    &   \cr
  &  1 &    &  4 &    &  6 &    &  4 &    &  1 &   \cr
1 &    &  5 &    & 10 &    & 10 &    &  5 &    & 1 \cr
\dots \cr
\end{array}\]
This fact is expressed formally in the identity:
\[\sum_{k=0}^{p}{k+n-1\choose n-1}  = {p+n\choose n} \]

Here is one simple way to prove the identity. First observe that ${k +
n-1 \choose n-1}$ is the number of ways of dividing $k$ objects into $n$
subsets: line up $k+n-1$ objects and select $n-1$ of them to mark the
boundaries of the $n$ subsets. The number of ways of choosing $n-1$ from
$k+n-1$ is, of course, $k+n-1\choose n-1$.  Now the sum on the left hand
side is the number of ways of dividing less than or equal to $p$ objects
into $n$ subsets, one term for each number of objects $k=0,\dots,p$.
The right hand side is the number of ways of dividing $p$ objects into
$n+1$ subsets.  By ignoring the first subest, every way of dividing $p$
objects into $n$ subsets gives exactly one way of dividing $k$ objects
into $n-1$ subsets where $k$ is $p$ minus the number of elements in
the ignored subset. Conversely, every way of dividing $k\le p$ objects
into $n$ subsets gives rise to exactly one way of dividing $p$ objects
into $n+1$ subsets: just add a subset with $p-k$ objects.  Therefore,
the two sides of the above identity must be equal.

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