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

What are the values for n and k?

My initial idea was to divide those two:



$$ \frac{k}{n+1-k}\:=\frac{2}{3} $$

After that sum them up:


And now devide the third with the first term and solve everything. However I would get both times $5k = 2n+2$ and that's the problem.

  • $\begingroup$ Well k is even and n = 4 mod 5 And you have $\frac {n!}{(k-1)!(n-k)!} = 1001 = 7.11.13$ You have enough for some real educated guesses. n = 14 k=6,works. $\endgroup$ – fleablood Nov 18 '15 at 19:10

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.

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    $\begingroup$ The part I like is that $$ \:\binom{n\:}{k-2}=1001.$$ I'm not sure that ever happens again, three consecutive entries $A,2A,3A$ in a row of Pascal's triangle $\endgroup$ – Will Jagy Nov 18 '15 at 18:51
  • $\begingroup$ @WillJagy that's an interesting question, actually :) $\endgroup$ – Servaes Nov 18 '15 at 18:53
  • $\begingroup$ @WillJagy That never happens again :( $\endgroup$ – Servaes Nov 18 '15 at 19:10
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    $\begingroup$ maybe not: this seems to have complete answer: math.stackexchange.com/questions/1495107/… $\endgroup$ – Will Jagy Nov 18 '15 at 19:23
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    $\begingroup$ My proof was about as short; dividing the binomial coefficients with ratios $1:2$ yields $n=3k-4$, and dividing those with rations $2:3$ yields $n=\tfrac52k-1$ as above. Hence $k=6$ and $n=14$. $\endgroup$ – Servaes Nov 18 '15 at 19:25

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

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  • $\begingroup$ Now try for the values $38116532895986727945334202400$ and $49378235797073715747364762200$ ;) $\endgroup$ – Servaes Nov 18 '15 at 19:02
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    $\begingroup$ @Servaes, give me a minute. $\endgroup$ – Will Jagy Nov 18 '15 at 19:05

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