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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.

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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
A related question on MO: – KCd Nov 25 '11 at 7:29
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 ( ) and tetrahedral numbers ( ) 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.

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

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

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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.

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

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Did you mean to write $\binom{10}{4} = \binom{21}{2}$ ? – Sasha Nov 25 '11 at 6:59
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
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


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