{ localUrl: '../page/factorial.html', arbitalUrl: 'https://arbital.com/p/factorial', rawJsonUrl: '../raw/5bv.json', likeableId: '3055', likeableType: 'page', myLikeValue: '0', likeCount: '6', dislikeCount: '0', likeScore: '6', individualLikes: [ 'EricBruylant', 'EmmanuelSmith', 'GuillermoAlcantara', 'KevinClancy', 'JaimeSevillaMolina', 'ThomasRichardson' ], pageId: 'factorial', edit: '4', editSummary: 'Pointed the induction links to the mathematical induction page', prevEdit: '3', currentEdit: '4', wasPublished: 'true', type: 'wiki', title: 'Factorial', clickbait: 'The number of ways you can order things. (Alternately subtitled: Is that exclamation point a factorial, or are you just excited to see me?)', textLength: '3408', alias: 'factorial', externalUrl: '', sortChildrenBy: 'likes', hasVote: 'false', voteType: '', votesAnonymous: 'false', editCreatorId: 'DouglasWeathers', editCreatedAt: '2016-07-17 18:54:44', pageCreatorId: 'MichaelCohen', pageCreatedAt: '2016-07-13 05:26:17', seeDomainId: '0', editDomainId: 'AlexeiAndreev', submitToDomainId: '0', isAutosave: 'false', isSnapshot: 'false', isLiveEdit: 'true', isMinorEdit: 'false', indirectTeacher: 'false', todoCount: '2', isEditorComment: 'false', isApprovedComment: 'true', isResolved: 'false', snapshotText: '', anchorContext: '', anchorText: '', anchorOffset: '0', mergedInto: '', isDeleted: 'false', viewCount: '228', 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%%', metaText: '', isTextLoaded: 'true', isSubscribedToDiscussion: 'false', isSubscribedToUser: 'false', isSubscribedAsMaintainer: 'false', discussionSubscriberCount: '2', maintainerCount: '2', userSubscriberCount: '0', lastVisit: '', hasDraft: 'false', votes: [], voteSummary: 'null', muVoteSummary: '0', voteScaling: '0', currentUserVote: '-2', voteCount: '0', lockedVoteType: '', maxEditEver: '0', redLinkCount: '0', lockedBy: '', lockedUntil: '', nextPageId: '', prevPageId: '', usedAsMastery: 'false', proposalEditNum: '0', permissions: { edit: { has: 'false', reason: 'You don't have domain permission to edit this page' }, proposeEdit: { has: 'true', reason: '' }, delete: { has: 'false', reason: 'You don't have domain permission to delete this page' }, comment: { has: 'false', reason: 'You can't comment in this 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