Find $n$ and $k$ if $\:\binom{n\:}{k-1}=2002\:\:\:\binom{n\:}{k}=3003\:\:$

Solution 1:

You have already found that $\frac{k}{n-k+1}=\frac23$ from which it follows that $$\frac32=\frac{n-k+1}{k}=\frac{n+1}{k}-1,$$ so $n=\tfrac52k-1$, which also shows that $k$ is a multiple of $2$ and that $n+1$ is a multiple of $5$. Let $k:=2m$ so that $n=5m-1$. Plugging this back in to $\tbinom{n+1}{k}=5005$ yields $$5005=\binom{n+1}{k}=\frac{(n+1)!}{(n-k+1)!k!}=\frac{(5m)!}{(3m)!(2m)!}.$$ You could note that $n\geq13$ because both $2002$ and $3003$ are multiples of $13$, and hence that $m\geq3$. Either way, trying small values of $m$ shows that $m=3$ works.

Solution 2:

I originally drew a large version of Pascal's Triangle after telling my father that the total number of gifts in The Twelve Days of Christmas song was $$ \:\binom{14\:}{3} = 364$$ and he complained, "What's wrong with $12?$" I tried to explain how the cumulative sums along the diagonals introduce a bit of an offset, The number of new gifts on day $n$ is $ \:\binom{n\:}{1},$ the total gifts on day $n$ is $ \:\binom{n + 1\:}{2},$ the total gifts on all days up to and including day $n$ is $ \:\binom{n + 2\:}{3}.$ He didn't like it. Later, I found a piece of paper where he had carefully written out the sums in $\Sigma$ notation, and got it right, of course.

Back to this problem, I have always liked that consecutive entries in row $14$ give $1001,2002,3003.$ I wonder whether that ever happens again, $A,2A,3A?$ Maybe not: this seems to give a complete answer: Find three consecutive entries of a row of Pascal triangle that are in the ratio of 1 : 2 : 3