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    "# [Combinatoric Selections](https://projecteuler.net/problem=53)\n",
    "\n",
    "Python/SageMath's [unlimited precision integers](https://docs.python.org/3/library/stdtypes.html#numeric-types-int-float-complex) make this trivial."
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       "4075"
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   "source": [
    "binoms = (binomial(n, r) for n in range(1, 101) for r in range(0, n + 1))\n",
    "len([b for b in binoms if b > 1000000])"
   ]
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   "source": [
    "If it weren't for unlimited precision, we would have a problem, since a lot of these [binomial coefficients](https://en.wikipedia.org/wiki/Binomial_coefficient) overflow a 64 bit integer - the largest one we encounter here is $\\binom{100}{50} = 100891344545564193334812497256$.\n",
    "\n",
    "However, even in that case there's a pretty simple workaround: finding $n$ and $r$ such that\n",
    "$$\\binom{n}{r} > 1000000$$\n",
    "is the same as finding $n$ and $r$ such that\n",
    "$$\\log{\\binom{n}{r}} > \\log{1000000}$$\n",
    "By applying the definition of the binomial coefficients and [logarithmic identities](https://en.wikipedia.org/wiki/List_of_logarithmic_identities), this turns to\n",
    "$$\\log{n!} - \\log{r!} - \\log{(n-r)!} > \\log{1000000}$$\n",
    "$\\log{100!} \\approx 363.739$, nowhere close to overflowing a float.\n",
    "\n",
    "$\\log{n!}$ can be computed in a number of ways. You can use the [log-gamma function](https://en.wikipedia.org/wiki/Gamma_function), or you can implement a function yourself, either by applying logarithmic identities again:\n",
    "$$\\log{n!} = \\log{1} + \\log{2} + \\log{3} \\cdots + \\log{n}$$\n",
    "or by using [Stirling's approximation](https://en.wikipedia.org/wiki/Stirling%27s_approximation):\n",
    "$$\\log{n!} \\approx \\left(n + \\frac{1}{2}\\right)\\log{n} - n + \\frac{1}{2}\\log{2\\pi}$$\n",
    "\n",
    "#### Copyright (C) 2025 filifa\n",
    "\n",
    "This work is licensed under the [Creative Commons Attribution-ShareAlike 4.0 International license](https://creativecommons.org/licenses/by-sa/4.0/) and the [BSD Zero Clause license](https://spdx.org/licenses/0BSD.html)."
   ]
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