Question invoving using binomial identities to determine $n$ and $k$ given $\binom{n}{k-1} = 2002$ and $\binom{n}{k} = 3003$ I have been trying to do a problem in a combinatorics textbook involving using binomial identities. The problem is :
"Determine $n,k \in \mathbb{N}$ from the equalities $\binom{n}{k-1} = 2002$ and
$\binom{n}{k} = 3003$"
The chapter that the problem is associated with has a lot of identities involving binomial coefficients. Some identities included are :

*

*$\binom{n}{k} = \binom{n}{n-k} \; 0 \leq k \leq n$

*$\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1} \; 1 \leq k \leq n$

*$\binom{n}{m} \binom{m}{k} = \binom{n}{k} \binom{n-k}{m-k} = \binom{n}{m-k}\binom{n-m+k}{k}$
Some other identities are also presented in examples.
I am not sure how to approach this problem. I can deduce that :
\begin{equation}
\frac{3}{2} = \frac{n-k+1}{k}
\end{equation}
given that :
\begin{equation}
\frac{3003}{2002} = \frac{1001}{1001} \frac{3}{2} = \frac{3}{2}
\end{equation}
and :
\begin{equation}
\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k}
\end{equation}
But I am not sure how to use this information and the identities to solve for $n$ and $k$.
Does anyone know how to approach a problem like this? There are no examples in the text. Maybe if I see how this problem can be solved then others like in the text will be easier.
 A: Starting from $$\frac{n-k+1}{k} = \frac{3}{2},$$ we obtain the Diophantine equation $$5k - 2n = 2$$ where $k < n \in \mathbb Z^+$.  Thus $k$ must be even, say  $k = 2m$, and the above equation becomes $$5m - n = 1,$$ or $n = 5m-1$, hence $$3003 = \binom{n}{k} = \binom{5m-1}{2m} = \frac{(5m-1)!}{(2m)!(3m-1)!}.$$  Since $3003 = 3 \cdot 7 \cdot 11 \cdot 13$, we require $5m-1 \ge 13$ or $m \ge 3$.  Since there are no larger prime factors, we also know $5m-1 < 17$, or $m \le 3$.  Therefore, $m = 3$ is the only candidate, and $$\binom{14}{6} = \frac{14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2} = \frac{(7 \cdot 13 \cdot 11 \cdot 3)(2 \cdot 12 \cdot 5 \cdot 2 \cdot 3)}{12 \cdot 5 \cdot 2 \cdot 2 \cdot 3} = 3003$$ as claimed.
A: First, note that $k$ is even. Thus, $k=2q$. Then
$$\frac32=\frac{n-2q+1}{2q}$$
Since the numerator must be $3q$, we know
$$n=3q+2q-1=5q-1$$
Thus
$$3003=\binom{n}{k}=\binom{5q-1}{2q}$$
Trying out different values of $q$ we find the solution $q=3$. This gives us $n=14$ and $k=6$.
A: Row 14 may be the only row with three consecutive entries in an arthmetic progression. I may have asked about that here or on MO, I'll check... NO, Marty Cohen asked about it Are there four consecutive binomial coefficients in a row in an arithmetic progression?
and Jack d'Aurizio came up with an infinite family of three-term arithmetic progressions in rows of Pascal's triangle, consecutive terms, beginning with 7,21,35 in row 7. OK, Jack's answer is at
Generalized Case: Three Consecutive Binomial Coefficients in AP

