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AlgebraPermutations and CombinationsPermutationsPermutations of all at a timeThe number of permutations of 𝑛 things taken all at a time: =𝑛(𝑛−1)(𝑛−2)⋯3⋅2⋅1≡𝑛! or 𝑛^{(𝑛)}ProofProof by Induction. Assume the formula to be true for 𝑛 things. Now take 𝑛+1 things. After each of these the remaining 𝑛 things may be arranged in 𝑛! ways, making in all 𝑛×𝑛!, that is (𝑛+1)!, permutations of 𝑛+1 things; therefore, ⋯.Permutations of 𝑟 thing at a timeThe number of permutations of 𝑛 things taken 𝑟 at a time is denoted by 𝑃(𝑛,𝑟). 𝑃(𝑛,𝑟)=𝑛(𝑛)(𝑛−1)(𝑛−2)⋯(𝑛−𝑟+1)≡𝑛^{(𝑟)}.ProofProof. By permutations of 𝑛 things; for (𝑛−𝑟) things are left out of each permutation; therefore 𝑃(𝑛,𝑟)=𝑛!÷(𝑛−𝑟)!. Observe that 𝑟=the number of factors.CombinationsCombinations of 𝑟 thing at a timeThe number of combinations of 𝑛 things taken 𝑟 at a time is denoted by 𝐶(𝑛,𝑟).𝐶(𝑛,𝑟)=
For every combination of 𝑟 things admits of 𝑟! permutations; therefore 𝐶(𝑛,𝑟)=𝑃(𝑛,𝑟)÷𝑟!.
𝐶(𝑛,𝑟) is greatest when 𝑟=12𝑛 or 12(𝑛±1) according as 𝑛 is even or odd. Homogeneous ProductsThe number of homogeneous products of 𝑟 dimensions of 𝑛 things is denoted by 𝐻(𝑛,𝑟). 𝐻(𝑛,𝑟)=𝑛(𝑛)(𝑛+1)(𝑛+2)⋯(𝑛+𝑟−1)1⋅2⋅3⋯𝑟≡ (𝑛+𝑟−1)^{(𝑟)}𝑟!When 𝑟 is >, this reduces to (𝑟+1)(𝑟+2)⋯(𝑛+𝑟−1)(𝑛−1)! Proof𝐻(𝑛,𝑟) is equal to the number of terms in the product of the expansions by the Binomial Theorem of the 𝑛 expressions (1−𝑎𝑥)^{−1}, (1−𝑏𝑥)^{−1}, (1−𝑐𝑥)^{−1}, ⋯. Put 𝑎=𝑏=𝑐=⋯=1. The number will be the coefficient of 𝑥^{𝑟} in (1−𝑥)^{−𝑛}.Permutations of alike thingsThe number of permutations of 𝑛 things taken all together, when 𝑎 of them are alike, 𝑏 of them alike, 𝑐 alike, ⋯. =𝑛!𝑎!𝑏!𝑐!⋯For, if the 𝑎 things were all different, they would form 𝑎! permutations where there is now but one. so of 𝑏, 𝑐, ⋯. Combinations of 𝑝 things foundThe number of combinations of 𝑛 things 𝑟 at a time, in which any 𝑝 of them will always be found, is =𝐶(𝑛−𝑝,𝑟−𝑝) For, if the 𝑝 things be set on one side, we have to add to them 𝑟−𝑝 things taken from the remaining 𝑛−𝑝 things in every possible way.Theorem of Combination𝐶(𝑛−1,𝑟−1)+𝐶(𝑛−1,𝑟)=𝐶(𝑛,𝑟) Proof by induction or as follows: Put one out of 𝑛 letters aside; there are 𝐶(𝑛−1,𝑟) combinations of the remaining 𝑛−1 letters 𝑟 at a time. To complete the total 𝐶(𝑛,𝑟), we must place with the excluded letter all the combinations of the remaining 𝑛−1 letters 𝑟−1 at a time.Combination of Different ThingsIf ther be one set of 𝑃 things, another of 𝑄 things, another of 𝑅 things, and so on; the number of combinations formed by taking one out of each set is =𝑃𝑄𝑅⋯, the product of the numbers in the several sets.For one of the 𝑃 things will form 𝑄 combinations with the 𝑄 things. A second of the 𝑃 things will form 𝑄 more combinations; ans so on. In all, 𝑃𝑄 combinations of two things. Similarly there will be 𝑃𝑄𝑅 combinations of three things; and so on. On the same principle, if 𝑝, 𝑞, 𝑟, ⋯ things be taken out of each set respectively, the number of combinations will be the product of the numbers of the separate combinations; that is, =𝐶(𝑃𝑝)⋅𝐶(𝑄𝑞)⋅𝐶(𝑅𝑟)⋅⋯ The number of combinations of 𝑛 things taken 𝑚 at a time, when 𝑝 of the 𝑛 things are alike, 𝑞 of them alike, 𝑟 of them alike, ⋯, will be the sum of all the combinations of each possible form of 𝑚 dimensions, and this is equal to the coefficient of 𝑥^{𝑚} in the expansion of (1+𝑥+𝑥^{2}+⋯+𝑥^{𝑝})(1+𝑥+𝑥^{2}+⋯+𝑥^{𝑞})(1+𝑥+𝑥^{2}+⋯+𝑥^{𝑟})⋯ The total number of possible combinations under the same circumstances, when the 𝑛 things are taken in all ways, 1, 2, 3, ⋯, 𝑛 at a time, =(𝑝+1)(𝑞+1)(𝑟+1)⋯−1 The number of permutations when they are taken 𝑚 at a time in all possible ways will be equal to the product of 𝑚! and the coefficient of 𝑥^{𝑚} in the expansion of 𝑥^{2}2!+ 𝑥^{3}3!+⋯+ 𝑥^{𝑝}𝑝! 𝑥^{2}2!+ 𝑥^{3}3!+⋯+ 𝑥^{𝑞}𝑞! Sources and Referenceshttps://archive.org/details/synopsisofelementaryresultsinpureandappliedmathematicspdfdrive©sideway ID: 210600008 Last Updated: 6/8/2021 Revision: 0 Ref: References
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