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International Journal of Science and Research (IJSR) ISSN: 2319-7064 Index Copernicus Value (2016): 79.57 | Impact Factor (2017): 7.296 Properties of Combination Using Double

G. Babu

Assistant Professor, Department of , Sri Akilandeshwari Women’s College Wandiwash

Abstract: In this paper contains to calculation some combination formulae using the techniques of double factorial. Applying this concept I have find out and proved the results of combination which had been already calculated.

Keywords: Combination, Factorial, Double factorial

1. Introduction Example: For n=10=2(5) The properties of combination is already proved using some 10‼ = 25 5! technique of factorial and sub factorial. In same concept I = 32 × 120 decided to verify and proved the properties of = 3840 using this double factorial. For n=9=2(5)-1 10! 9‼ = 2. Basic Definitions 25 5! 3628800 = 2.1 Definition of Combination 3840 = 945 A permutation is an arrangement of all or part of a of objects, where the order doesn’t matter. 3.1 Useful theorem

Computing the number of combination of r objects chosen n! n! For any non negative integer n, then = n − 1 ‼ (or) from n objects is nCr = n!! r! n−r ! n! = n − 1 !! × n!!

2.2 Definition of factorial Proof Factorial says to multiply all whole numbers from the 1) If n is odd chosen number down to 1. The symbol is “! " The formula is n! n n − 1 n − 2 … … × 3 × 2 × 1 n! = n n − 1 n − 2 … … . .× 3 × 2 × 1 = n‼ n n − 2 n − 4 … … × 5 × 3 × 1 2.3 Definition of double factorial = n − 1 n − 3 … … × 4 × 2 = (n − 1)‼ The double factorial or semi factorial of a number n(denoted by n!! ) is the product of all the integers from 1 up to n that 2) If n is even n! n n − 1 n − 2 … … × 3 × 2 × 1 have the same parity (odd or even) as n. Formula is = n‼ n n − 2 n − 4 … … × 6 × 4 × 2 For n is even, then = n − 1 n − 3 … … × 3 × 1 n‼ = n n − 2 n − 4 … … × 6 × 4 × 2 = (n − 1)‼

For n is odd , then Combining this cases for any non negative integer n n! n‼ = n n − 2 n − 4 … … × 5 × 3 × 1 = n − 1 ‼ n!! Example: ⇔ n! = n − 1 !! × n!! 10‼ = 10 × 8 × 6 × 4 × 2 = 3840 Example 9‼ = 9 × 7 × 5 × 3 × 1 For n= 6 = 945 n! = 6! Note : As n! ! and not n!! = 6 × 5 × 4 × 3 × 2 × 1 = 720 3. Relation between factorial and double n − 1 ‼ × n‼ = 5‼ × 6‼ factorial = 5 × 3 × 1 × 6 × 4 × 2 = 720

For even n=2x, x≥ 0 then x Result 1 n‼ = 2 x! 0‼ = 1 and −1 ‼ = 1 For odd n=2x-1 , x≥ 1 then (2x)! n‼ = Proof 2x x! w.k.t n‼ = 2x x!

Volume 7 Issue 12, December 2018 www.ijsr.net Licensed Under Creative Commons Attribution CC BY Paper ID: ART20193805 10.21275/ART20193805 1183 International Journal of Science and Research (IJSR) ISSN: 2319-7064 Index Copernicus Value (2016): 79.57 | Impact Factor (2017): 7.296 0‼ = 20 0! Result 6 = 1 nCr = nCp then r = p or r + p = n

Result 2 Proof: n‼ = n n − 2 ‼ n + 1 ‼ = (n + 1) n − 1 ‼

Example: 6‼ = 6 × 4‼ Equating (1) and (2) = 6 × 4 × 2‼ r‼ = p‼ then r=p or r+p=n = 6 × 4 × 2

= 48 Example:

6‼ × 5 ‼ Result 3 6C1 = n‼ × (n − 1)‼ 1‼ × 5 ‼ 0 ‼ × 4 ‼ nCr = 6 × 4 × 2 r‼ × (r − 1)‼ n − r − 1 ‼ × n − r ‼ = 4 × 2 = 6 Proof: n! 6‼ × 5 ‼ nC = 6C5 = r r! n − r ! 5‼ × 1 ‼ 4 ‼ × 0 ‼ 6 × 4 × 2 Since, n! = n − 1 !! × n!! = Therefore, 4 × 2 n‼ × (n − 1)‼ = 6 nCr = r‼ × (r − 1)‼ n − r − 1 ‼ × n − r ‼ Result 7 np nC = r Example: r r! 5‼ × 4‼ 5C = 2 2‼ × (1)‼ 2 ‼ × 3 ‼ Proof: 5 × 3 × 1 × 4 × 2 n‼ × n − 1 ‼ = nC = 2 × 1 × 2 × 3 r r‼ × r − 1 ‼ n − r − 1 ‼ × n − r ‼ = 10 np = r r‼ × r − 1 ‼ Result 4 npr = nC0 = nCn = 1 r!

Proof: Example: n‼ × n − 1 ‼ 6‼ × 5‼ nC = 6C3 = r r‼ × r − 1 ‼ n − r − 1 ‼ × n − r ‼ 3‼ × 2‼ × 2‼ × 3‼ n‼ × n − 1 ‼ 6 × 4 × 2 × 5 × 3 nC = = 0 0‼ × −1 ‼ n − 1 ‼ × n ‼ 3 × 2 × 2 × 3 = 20 = 1

n‼ × n − 1 ‼ 6‼ × 5‼ nCn = n‼ × n − 1 ‼ −1 ‼ × 0 ‼ 6p3/3! = / 3! 2‼ × 3‼ = 1 6 × 4 × 2 × 5 × 3 × 1 = /6 2 × 3 × 1 Result 5 = 120/6 nC1 = n =20 Proof: n‼ × n − 1 ‼ Result 8 nC1 = 1‼ × 0 ‼ n − 2 ‼ × n − 1 ‼ nCr + nCr−1 = (n + 1)Cr n n − 2 ‼ = 1‼ × 0 ‼ n − 2 ‼ Proof: = n w.k.t (n + 1)Cr Example n + 1 ‼ × n ‼ = ______(1) 6‼ × 5 ‼ r‼ × r − 1 ‼ n − r + 1 ‼ × n − r ‼ 6C1 = 1‼ × 5 ‼ 0 ‼ × 4 ‼ LHS=nCr + nCr−1 6 × 4 × 2 = 4 × 2 = 6

Volume 7 Issue 12, December 2018 www.ijsr.net Licensed Under Creative Commons Attribution CC BY Paper ID: ART20193805 10.21275/ART20193805 1184 International Journal of Science and Research (IJSR) ISSN: 2319-7064 Index Copernicus Value (2016): 79.57 | Impact Factor (2017): 7.296 n‼ × n − 1 ‼ n‼ × n + 1 ‼ 1 = = r‼ × r − 1 ‼ n − r − 1 ‼ × n − r ‼ (n + 1) × r − 1 ‼ × n − r ‼ r‼ × n − r − 1 ‼ n‼ × n − 1 ‼ 1 + + r − 2 ‼ × r − 1 ‼ n − r + 1 ‼ × n − r ‼ r − 2 ‼ × n − r + 1 ‼

(n + 1)n‼ × n − 1 ‼ = (n + 1)r‼ × r − 1 ‼ n − r − 1 ‼ × n − r ‼ (n + 1)n‼ × n − 1 ‼ + (n + 1) r − 2 ‼ × r − 1 ‼ n − r + 1 ‼ × n − r ‼

n‼ × n + 1 ‼ n − r + 1 n − r − 1 ‼ r − 2 ‼ + r r − 2 ‼ n − r − 1 ‼ = n + 1 × r − 1 ‼ × n − r ‼ r‼ × n − r − 1 ‼ r − 2 ‼ × n − r + 1 ‼

n‼ × n + 1 ‼ n − r − 1 ‼ r − 2 ‼ n − r + 1 +r = n + 1 × r − 1 ‼ × n − r ‼ r‼ × n − r − 1 ‼ r − 2 ‼ × n − r + 1 ‼

n‼ × n + 1 ‼ n + 1 = 2n = n + 1 × r − 1 ‼ × n − r ‼ r‼ n − r + 1 ‼ Since using binomial formula n n − 1 n + 1 ‼ × n ‼ (a + b)n = an + nan−1b1 + an−2b2 + ⋯ + bn = 2! r‼ × r − 1 ‼ n − r + 1 ‼ × n − r ‼ We observe this a=1, b=1

= n + 1 Cr by 1 Result 10 nC + nC + nC + ⋯ + ⋯ = nC + nC + nC + ⋯ + ⋯ Example: 0 2 4 1 3 5 = 2n−1 nCr + nCr−1 = (n + 1)Cr

Where n=5, r=2 Proof: nC0 + nC2 + nC4 + ⋯ + ⋯ LHS=5C2 + 5C1 5‼ × 4 ‼ 5‼ × 4 ‼ n n − 1 = + = 1 + 2‼ × 3 ‼ 1 ‼ × 2 ‼ 1‼ × 4 ‼ 0 ‼ × 3 ‼ 2! 4 × 2 × 5 × 3 4 × 2 × 5 × 3 n n − 1 n − 2 n − 3 = + + + ⋯ 3 × 2 × 2 3 × 4 × 2 4! = 10 + 5 = 15 1 n n 6‼ × 5 ‼ = (1 + 1) + 0 RHS = 6C = 2 2 2‼ × 4 ‼ 1 ‼ × 3 ‼ = 2n−1 6 × 4 × 2 × 5 × 3 = nC1 + nC3 + nC5 + ⋯ + ⋯ 4 × 2 × 2 × 3 n n − 1 n − 2 = 15 = n + 3! n n − 1 n − 2 n − 3 n − 4 Result 9 + + ⋯ n 5! nC0 + nC1 + nC2 + ⋯ + nCn = 2 1 = (1 + 1)n + 0n 2 Proof = 2n−1 w.k.t n‼ × n − 1 ‼ nC0 = = 1 Since binomial formula 0‼ × −1 ‼ n − 1 ‼ × n ‼ n n n‼ × n − 1 ‼ a + b + a − b nC = = n 1 1‼ × 0 ‼ n − 2 ‼ × n − 1 ‼ 2

n‼ × n − 1 ‼ n n − 1 n n n − 1 n−2 2 nC = = = a + a b 2 2‼ × 1 ‼ n − 3 ‼ × n − 2 ‼ 2! 2!

n‼ × n − 1 ‼ n n − 1 (n − 2) n n − 1 n − 2 n − 3 n−4 4 nC = = + a b + ⋯ 3 3‼ × 2 ‼ n − 3 ‼ × n − 4 ‼ 3! 4! n n Etc… a + b − a − b n‼ × n − 1 ‼ n n − 1 2 nC = = n−2 2‼ × 1 ‼ n − 3 ‼ × n − 2 ‼ 2! n n − 1 n − 2 = nan−1b1 + an−3b3 n‼ × n − 1 ‼ 3! nCn−1 = = n 1‼ × 0 ‼ n − 2 ‼ × n − 1 ‼ n n − 1 n − 2 n − 3 n − 4 n−5 5 n‼ × n − 1 ‼ + a b nC = = 1 5! n 0‼ × −1 ‼ n − 1 ‼ × n ‼ + ⋯

4. Conclusion n = 1 + 1 Volume 7 Issue 12, December 2018 www.ijsr.net Licensed Under Creative Commons Attribution CC BY Paper ID: ART20193805 10.21275/ART20193805 1185 International Journal of Science and Research (IJSR) ISSN: 2319-7064 Index Copernicus Value (2016): 79.57 | Impact Factor (2017): 7.296 In this paper I have find out and proved some results with example, properties of combinations applying the technique of double factorial. It is used in many areas. Already I have calculated and proved some results with example, properties of permutation applying the technique of double factorial and I extend this paper to solve the same concept to find out the properties of combinations using double factorial.

References

[1] Ramesh Chandra, “Permutation and combinations” book. [2] Anglani R, Barlie M. as Sums. arXiv:Math/0702010v1 [Math.HO] 2007. [3] Bhargava M. The Factorial Function and Generalizations. Math. Assoc. America. Am. Math. Monthly. 107:783–799. 2000. [4] Culham JR. Factorial, Gamma and Beta Functions. Engineering Course ME755- Special Functions. University of Waterloo, Waterloo. 2014. [5] Dartmouth College . Correspondance Between Leonhard Euler and Chr. Goldbach, 1729–1763. 2014. pp. 1–59. [6] Davis PJ. Leonhard Euler’s Integral: A Historical Profile of the Gamma Function. In Memorium: Milton Abramowitz. Math. Assoc. America. Amer Math Monthly. 66:849–869. 1959.

Volume 7 Issue 12, December 2018 www.ijsr.net Licensed Under Creative Commons Attribution CC BY Paper ID: ART20193805 10.21275/ART20193805 1186