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text: 'Factorial is most simply defined as a [-3jy] on positive [48l integers]. 5 factorial (written as $5!$) means $1*2*3*4*5$. In general then, for a positive integer $n$, $n!=\\prod_{i=1}^{n}i$. For applications to [ combinatorics], it will also be useful to define $0! = 1$.\n\n## Applications to Combinatorics ##\n\n$n!$ is the number of possible orders for a set of $n$ objects. For example, if we arrange the letters $A$, $B$, and $C$, here are all the options:\n$$ABC$$\n$$ACB$$\n$$BAC$$\n$$BCA$$\n$$CAB$$\n$$CBA$$\nYou can see that there are $6$ possible orders for $3$ objects, and $6 = 3*2*1 = 3!$. Why does this work? We can [5fz prove this by induction]. First, we'll see pretty easily that it works for $1$ object, and then we can show that if it works for $n$ objects, it will work for $n+1$. Here's the case for $1$ object.\n$$A$$\n$$1 = \\prod_{i=1}^{1}i = 1!$$\nNow we have the objects $\\{A_{1},A_{2},...,A_{n},A_{n+1}\\}$, and $n+1$ slots to put them in. If $A_{n+1}$ is in the first slot, now we're ordering $n$ remaining objects in $n$ remaining slots, and by our [5fz induction hypothesis], there are $n!$ ways to do this. Now let's suppose $A_{n+1}$ is in the second slot. Any orderings that result from this will be completely unique from the orderings where $A_{n+1}$ was in the first slot. Again, there are $n$ remaining slots, and $n$ remaining objects to put in them, in an arbitrary order. There are another $n!$ possible orderings. We can put $A_{n+1}$ in each slot, one by one, and generate another $n!$ orderings, all of which are unique, and by the end, we will have every possible ordering. We know we haven't missed any because $A_{n+1}$ has to be somewhere. The total number of orderings we get is $n!*(n+1)$, which equals $(n+1)!$.\n\n\n## Extrapolating to [50d Real Numbers] ##\n\nThe factorial function can be defined in a different way so that it is defined for all real numbers (and in fact for complex numbers too).\n\n%%hidden(Definition):\nWe define $x!$ as follows:\n$$x! = \\Gamma (x+1),$$\nwhere $\\Gamma $ is the [ gamma function]:\n$$\\Gamma(x)=\\int_{0}^{\\infty}t^{x-1}e^{-t}\\mathrm{d} t$$\nWhy does this correspond to the factorial function as defined previously? We can prove by induction that for all positive integers $x$:\n$$\\prod_{i=1}^{x}i = \\int_{0}^{\\infty}t^{x}e^{-t}\\mathrm{d} t$$\nFirst, we verify for the case where $x=1$. Indeed:\n$$\\prod_{i=1}^{1}i = \\int_{0}^{\\infty}t^{1}e^{-t}\\mathrm{d} t$$\n$$1=1$$\nNow we suppose that the equality holds for a given $x$:\n$$\\prod_{i=1}^{x}i = \\int_{0}^{\\infty}t^{x}e^{-t}\\mathrm{d} t$$\nand try to prove that it holds for $x + 1$:\n$$\\prod_{i=1}^{x+1}i = \\int_{0}^{\\infty}t^{x+1}e^{-t}\\mathrm{d} t$$\nWe'll start with the induction hypothesis, and manipulate until we get the equality for $x+1$.\n$$\\prod_{i=1}^{x}i = \\int_{0}^{\\infty}t^{x}e^{-t}\\mathrm{d} t$$\n$$(x+1)\\prod_{i=1}^{x}i = (x+1)\\int_{0}^{\\infty}t^{x}e^{-t}\\mathrm{d} t$$\n$$\\prod_{i=1}^{x+1}i = (x+1)\\int_{0}^{\\infty}t^{x}e^{-t}\\mathrm{d} t$$\n$$= 0+\\int_{0}^{\\infty}(x+1)t^{x}e^{-t}\\mathrm{d} t$$\n$$= \\left (-t^{x+1}e^{-t}) \\right]_{0}^{\\infty}+\\int_{0}^{\\infty}(x+1)t^{x}e^{-t}\\mathrm{d} t$$\n$$= \\left (-t^{x+1}e^{-t}) \\right]_{0}^{\\infty}-\\int_{0}^{\\infty}(x+1)t^{x}(-e^{-t})\\mathrm{d} t$$\nBy the product rule of integration:\n$$=\\int_{0}^{\\infty}t^{x+1}e^{-t}\\mathrm{d} t$$\nThis completes the proof by induction, and that's why we can define factorials in terms of the gamma function.\n%%',
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