Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $\binom{n}{k}=\binom{n'}{k'}$ with $k \geq 2$, $k' \geq 2$, $n \geq 2k$ and $n' \geq 2k'$. Does it follow that $n=n'$ and $k=k'$?

EDIT: Yup, $\binom{16}{2}=\binom{10}{3}=120$.

Now I want to ask if there are infinitely many such pairs, but I should probably ask that in a separate question. Thanks!

EDIT 2: for future reference, yes there are infinitely many such coincidences. See Singmaster's Conjecture.

share|improve this question
    
Rephrased: "Are there numbers in Pascal's triangle that are equal, apart from the obvious right-left symmetry?" I haven't checked, but I am convinced that there are quite a few. –  Arthur Nov 25 '11 at 7:01
3  
A related question on MO: mathoverflow.net/questions/17058/factorials-in-pascals-triangle –  KCd Nov 25 '11 at 7:29

5 Answers 5

up vote 3 down vote accepted

There are one or two (nontrivial) choices of $(m,n)$ where $\binom{n}{2} = \binom{m}{3}$. Comparing the sequences on OEIS might be faster than coming up with the correct words to find it in a search engine.

Edit: 120 is in both sequences, with $m=10$ and $n=16$. In OEIS the list of triangular numbers ( http://oeis.org/A000217 ) and tetrahedral numbers ( http://oeis.org/A000292 ) shows that there are no other small examples and I am pretty sure that this pair is known to be the largest integer solution of $3y(y-1)=x(x-1)(x-2)$, and that this was first proved in an elementary way that does not use the theory of elliptic curves.

share|improve this answer

Let $f(t)$ be the number of ways in which $t$ can be written as a binomial coefficient. Then the question can be restated equivalently as asking whether $f(t) \leqslant 2$ for all $t$. As the other answers point out, there are numerous counterexamples to this strong claim.

It is however likely that a weaker form of this statement is true. Singmaster's conjecture states that there exists some $M \lt \infty$ such that $f(t) \leqslant M$ for all $t$. This is still open. The best known upper bound, due to D. Kane, is $$f(t) = O \left(\frac{(\log t) \cdot (\log \log \log t) }{(\log \log t)^3} \right) .$$ On the other hand, it is consistent with our current knowledge that $f(t)$ never exceeds $8$.

Reference: This MO post by Michael Hardy asks for recent progress on this problem.

share|improve this answer

$$ \binom{3003}{1} = \binom{78}{2} = \binom{15}{5} = \binom{14}{6}. $$

Nobody knows whether any number besides 3003 appears at least eight times in Pascal's triangle.

share|improve this answer

10-choose-4 = 21-choose-2.${}{}{}{}$

share|improve this answer
    
Did you mean to write $\binom{10}{4} = \binom{21}{2}$ ? –  Sasha Nov 25 '11 at 6:59
1  
Trying to guess what you meant, since it obviously isn't $\binom{10}{2}=\binom{21}{4}$... –  Alon Amit Nov 25 '11 at 7:01
    
${10\choose 4} = {21 \choose 2}$ –  Chris Taylor Nov 25 '11 at 7:02
1  
Oops. Miscopied from my source. Thanks for spotting the error and figuring out what I meant. –  Gerry Myerson Nov 25 '11 at 10:13
    
@Gerry I took the liberty to correct the typo. Hope it's ok. Regards, –  Srivatsan Nov 25 '11 at 13:42

$$\binom{10}{3}=\binom{16}{2}.$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.