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I've just seen a gift card on the internet that is supposedly for a mathematician's 21st birthday. It says

$$ \text{Happy } ^{10} C_3 - 11\ln(e)-\frac{289}{3}+\left(\int_{\pi/6}^{\pi/3}\sec^2(x)dx\right)^2+7$$

I don't think I have ever seen the notation $^{10} C_3$ before. What does it mean?

Well,

$$ \left(\int_{\pi/6}^{\pi/3}\sec^2(x)dx\right)^2= \left(\tan(\pi/3)-\tan(\pi/6)\right)^2=\left(\sqrt{3}-\sqrt{1/3}\right)^2=4/3,$$

so I can simplify the expression that is wished to be happy to

$$ ^{10} C_3 - 11-\frac{289}{3}+4/3+7 = {}^{10} C_3 -99.$$

I guess the result is supposed to be 21, so (if I have not made any mistakes) I would expect that $${}^{10} C_3=120,$$ but I have absolutely no idea how to make sense of this...

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    $\begingroup$ $10 $ choose $3= \dfrac{10!}{7!\, 3!}=120$ $\endgroup$ – J. W. Tanner Jul 4 at 0:08
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    $\begingroup$ $^nC_r$ is an alternate notation for the binomial symbol $\binom nr=\frac {n!}{r!(n-r)!}$. $\endgroup$ – lulu Jul 4 at 0:09
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${}^{10} C_3$ means "$10$ choose $3$" and is more frequently denoted by the binomial coefficient $\binom{10}{3}$.

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    $\begingroup$ See also en.wikipedia.org/wiki/Binomial_coefficient#History_and_notation $\endgroup$ – lhf Jul 4 at 0:09
  • $\begingroup$ Thanks! That was easy... Well, of course I know the binomial coefficient, but I have never seen that notation before. Is it commonly used (at least in English-speaking countries)? $\endgroup$ – Mars Plastic Jul 4 at 0:11
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    $\begingroup$ Instead of "more frequently" say "in more advanced math". That ${}^nC_r$ notation is probably in common use in the US for ages 12 to 18 or so. Along with a corresponding ${}^nP_r$ notation. Called "combinations and permutations". $\endgroup$ – GEdgar Jul 4 at 0:12
  • $\begingroup$ ... Here is an example: britannica.com/science/permutation $\endgroup$ – GEdgar Jul 4 at 0:18
  • $\begingroup$ though in the Britannica example, $n$ is subscript rather than superscript, @GEdgar $\endgroup$ – J. W. Tanner Jul 4 at 0:20
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$$^{10}C_3 = {10 \choose 3} = \cfrac{10!}{7! 3!} = \cfrac{ \Gamma(11)}{\Gamma(8) \Gamma(4)} = 120 $$ You’ve guessed the right value.

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