Combinations of n elements in pairs

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  • Thread starter archaic
  • Start date
  • #1
642
196
Hey!
[tex]
C^2_2 = 1\:\:\:C^2_3 = 3\:\:\:C^2_4 = 6\:\:\:C^2_5 = 10 \\
We\:can\:see\:that\:\:\:C^2_n = C^2_{n-1} + (n-1),\:let's\:try\:n = 5 \\
C^2_5 = C^2_4 + (5 - 1) \\
= C^2_3 + (4 - 1) + (5 - 1) \\
= C^2_2 + (3 - 1) + (4 - 1) + (5 - 1) \\
= 1 + (3 - 1) + (4 - 1) + (5 - 1) \\
C^2_5 = (1 - 1) + (2 - 1) + (3 - 1) + (4 - 1) + (5 - 1) = 10 \\
Generally,\: C^2_n = (1 - 1) + (2 - 1)\:+\:...\:+\:(n - 1) \\
= (1 + 2\: +\: ... \: + \: n) - n \\
= \frac{(n+1)n}{2} - n \\
= \frac{n^2-n}{2}
[/tex]
 

Answers and Replies

  • #2
mathman
Science Advisor
7,869
450
What is the point? Standard combinatorial formula [tex]\binom{n}{k}=\frac{n!}{k!(n-k)!}[/tex].
 
  • #3
642
196
What is the point? Standard combinatorial formula [tex]\binom{n}{k}=\frac{n!}{k!(n-k)!}[/tex].
well, call it a highschooler's curiosity and then impulse
 
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