I'm stuck on this one. I am not expected to know how to solve cubic equations so this gets even more confusing, as i get $\binom n 3 = 120$. So I can't even calculate $n$.
Is there a way to go around solving the cubic equation?
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1You can solve $\binom{n}{3} = 120$ by trial and error, no theory needed. – Daniel Fischer May 23 '15 at 21:19
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$n(n-1)(n-2)=6\times 120$ so take the cube root of $720$ to get an estimate. Do you know a cube near $720$? – Mark Bennet May 23 '15 at 21:21
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There do exist ways of solving cubic equations explicitly (as well as solving quartics). It is of interest to note that the quintic however is not always solvable (one of the big results from Galois Theory). The cubic can be solved via Cardano's Formula – JMoravitz May 23 '15 at 21:24
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Very true, Daniel, it seems I cop out too early on most of these questions. Thanks all. – John Doe May 23 '15 at 21:27
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We have $$\dfrac{n(n-1)(n-2)}{3!} = 120 \implies (n-1)^3-(n-1) = 720 \implies (n-1)^3 > 720$$ Hence, $n > 1+\sqrt[3]{720}$. By a same argument $(n-2)^3 < 720$, which implies that $n < 2+\sqrt[3]{720}$. It is not that hard to approximate $\sqrt[3]{720}$, since $9^3 = 729$. This means $9<n<11$. Hence, the only possible integer is $10$, which indeed satisfied $\dbinom{10}3 = 720$.
Adhvaitha
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