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Total Possible Arrangement


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Total Number of Possible Arrangement
 Pascal's Formula
 Binomial Expansion
 Pascal's Triangle
 Multinomial Expansion
 Number of Subsets of a Set
 Total Possible Arrangements
 Finite Differences

Total Number of Possible Arrangement

Pascal's Formula

Pascal's Formula, also called Pascal's Rule is a combinatiorial identity. Definition (Pascal's Formula) For the 𝑟-elemnet subsets of an 𝑛 element set, the following combinatorial relationship holds: 𝐶(𝑛,𝑟)=𝐶(𝑛−1,𝑟)+𝐶(𝑛−1,𝑟−1) Proof 𝐶(𝑛,𝑟)=𝑛(𝑛−𝑟)!𝑟!  =(𝑛−1)!𝑛(𝑛−𝑟)!𝑟!  =(𝑛−1)!(𝑛−𝑟)(𝑛−𝑟)!𝑟!+(𝑛−1)!𝑟(𝑛−𝑟)!𝑟!  =(𝑛−1)!(𝑛−𝑟−1)!𝑟!+(𝑛−1)!(𝑛−𝑟)!(𝑟−1)!  =(𝑛−1)!((𝑛−1)−𝑟)!𝑟!+(𝑛−1)!((𝑛−1)−(𝑟−1)!(𝑟−1)!  =𝐶(𝑛−1,𝑟)+𝐶(𝑛−1,𝑟−1)

Binomial Expansion

Binomial Expansion is an algebraic expressions of two terms. Binomial ExpansionThe expansion of the binomial expression (𝑎+𝑏)𝑛 is (𝑎+𝑏)𝑛=𝐶(𝑛,𝑛)𝑎𝑛+𝐶(𝑛,𝑛−1)𝑎𝑛−1𝑏+𝐶(𝑛,𝑛−2)𝑎𝑛−2𝑏2+⋯+𝐶(𝑛,1)𝑎𝑏𝑛−1+𝐶(𝑛,0)𝑎𝑏𝑛

Pascal's Triangle

Pascal's triangle is a triangular array of the binomial coefficients.

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Multinomial Expansion

Binomials are just a special case of a larger class of expressions called multinomials expressions with more than one term. The expresion (𝑎+𝑏+𝑐) is a trinomial.

Number of Subsets of a Set

For a 10 elements set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, there are 2^10 possible sets. In mathematics of sets, This is not a proper subset of the original set, because it contains the entire set. All other subsets, including the empty set, are considered proper subsets. Therefore, there are 2^10−1 proper subsets of {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. More generally, there are 2^𝑛 subsets of an 𝑛-elemnet set, and 2^𝑛-1 proper subsets of that 𝑛n-element set.

Total Possible Arrangements

Example of a binomial birth-order arrangements. For a family with 𝑛 children, 𝑟 of them sons, there are 𝐶(𝑛,𝑟) different birth order arrangements. 𝑛𝑟=0𝐶(𝑛,𝑟)=𝐶(𝑛,0)+𝐶(𝑛,1)+⋯+𝐶(𝑛,𝑟)+⋯+𝐶(𝑛,𝑛)

The problem of total possible arrangements can be solved by relating the combinatorial representation to the binomial expransion. (𝑎+𝑏)𝑛=𝐶(𝑛,𝑛)𝑎𝑛+𝐶(𝑛,𝑛−1)𝑎𝑛−1𝑏+𝐶(𝑛,𝑛−2)𝑎𝑛−2𝑏2+⋯+𝐶(𝑛,1)𝑎𝑏𝑛−1+𝐶(𝑛,0)𝑎𝑏𝑛

Two equations can be equated by letting 𝑎 and 𝑏 both equal 1, 2𝑛=𝐶(𝑛,𝑛)+𝐶(𝑛,𝑛−1)+𝐶(𝑛,𝑛−2)+⋯+𝐶(𝑛,1)+𝐶(𝑛,0)

Therefore 𝑛𝑟=0𝐶(𝑛,𝑟)=2𝑛 And the sum of the entries in the 𝑛th row of Pascal's triangle is 2𝑛.

Total Possible ArrangementsThe total number of possible ways to arrange 𝑛 objects with first type of object from 0 to 𝑛 and second type of object from 𝑛 to 0 is 𝑛𝑟=0𝐶(𝑛,𝑟)=𝐶(𝑛,0)+𝐶(𝑛,1)+⋯+𝐶(𝑛,𝑟)+⋯+𝐶(𝑛,𝑛)=2𝑛 Or Total Possible ArrangementsGiven 𝑛 objects of 𝑟 different types, then there are 𝑟𝑛 total possible ways to arrange the 𝑛 objects given all possible ways to group the 𝑟 types.

Finite Differences

Pascal's Triangle


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References

  1. B. Joseph, 1978, University Mathematics: A Textbook for Students of Science & Engineering, Blackie & Son Limited, HongKong
  2. Wheatstone, C., 1854, On the Formation of Powers from Arithmetical Progressions, Proceedings of The Royal Society of London, Vol 7, p145-151,, London
  3. Stroud, K.A., 2001, Engineering Mathematics, Industrial Press, Inc, NY
  4. Coolidge, J.L., 1949, The Story of The Binomial Theorem, The American Mathematical Monthly, Vol 56, No.3, Mar, pp147-157
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ID: 190500011 Last Updated: 2019/5/11 Revision: Ref:

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