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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.

Kenta S
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scipio
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    the neat thing is that, with $n=14,$ three numbers in a row are $1001, 2002, 3003.$ I'll post Pascal's triangle – Will Jagy Jul 18 '20 at 01:43
  • $\frac{3}{2} = \frac{n - k + 1}{k}$ gives you $k = \frac{2}{5} (n + 1)$, so $n$ must be $4 \pmod 5$. – Toby Mak Jul 18 '20 at 01:50
  • $3003$ is the only number other than $1$ that is known to occur as many as eight times in Pascal's triangle. Whether any other number appears so many times, or more than that many, is unknown. Singmaster's conjecture says there is a finite upper bound. As far as empirical evidence known so far tells us, the smallest upper bound may be $8.$ $$ \binom{3003} 1 = \binom{78} 2 = \binom{15} 5 = \binom{14} 6 = \binom{14} 8 = \binom {15}{10} = \binom{78}{76} = \binom{3003}{3002}. $$ – Michael Hardy Jul 18 '20 at 03:39

3 Answers3

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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.

heropup
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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 $k=6$.

QC_QAOA
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  • How did you prove that $k$ must be even? – Toby Mak Jul 18 '20 at 01:58
  • The denominator must be a multiple of $2$. Otherwise it wouldn’t simplify to $\frac32$. – QC_QAOA Jul 18 '20 at 02:00
  • $n$ choose $k-1$ can be written as $\frac{n!}{(k-1)! (n-k+1)!}$. However, both $k = 4 \Rightarrow (k-1)! = 3! = 3 \cdot 2 \cdot 1$ and $k = 3 \Rightarrow (k - 1)! = 2! = 2 \cdot 1$ give an even denominator. – Toby Mak Jul 18 '20 at 02:05
  • Why is it the only solution? – markvs Jul 18 '20 at 02:05
  • Using Stirling’s Approximation it is easy to show that $\binom{5q-1}{2q}$ is increasing – QC_QAOA Jul 18 '20 at 02:13
  • @Toby_Mak The question is not about the denominator of the binomial in its unsimplified form (the denominator will always be $1$ since the binomials are integers), it’s about the ratio between the two binomials specified in the question – QC_QAOA Jul 18 '20 at 02:20
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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

enter image description here

Will Jagy
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    Row $7$ also has an arithmetic progression with $7, 21, 35$. Checking up to row $100$, rows $7, 14, 23, 34, 47, 62, 79, 98$ have an arithmetic progression. – Varun Vejalla Jul 18 '20 at 01:58
  • @VarunVejalla thank you. I put in a link, Jack's infinite family is in a comment there. – Will Jagy Jul 18 '20 at 01:59
  • @VarunVejalla Jack answered at https://math.stackexchange.com/questions/1018687/generalized-case-three-consecutive-binomial-coefficients-in-ap – Will Jagy Jul 18 '20 at 02:02
  • Neat! I was just about to suggest that these followed $n^2 - 2$ after looking at the OEIS (which is confirmed by the answer you sent). – Varun Vejalla Jul 18 '20 at 02:06
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    The question was about two entries of the triangle, not three, – markvs Jul 18 '20 at 02:06