Combinations of n elements in pairs

[archaic]

Hey!
$$C^2_2 = 1\:\:\:C^2_3 = 3\:\:\:C^2_4 = 6\:\:\:C^2_5 = 10 \\<br /> We\:can\:see\:that\:\:\:C^2_n = C^2_{n-1} + (n-1),\:let's\:try\:n = 5 \\<br /> C^2_5 = C^2_4 + (5 - 1) \\<br /> = C^2_3 + (4 - 1) + (5 - 1) \\<br /> = C^2_2 + (3 - 1) + (4 - 1) + (5 - 1) \\<br /> = 1 + (3 - 1) + (4 - 1) + (5 - 1) \\<br /> C^2_5 = (1 - 1) + (2 - 1) + (3 - 1) + (4 - 1) + (5 - 1) = 10 \\<br /> Generally,\: C^2_n = (1 - 1) + (2 - 1)\:+\:...\:+\:(n - 1) \\<br /> = (1 + 2\: +\: ... \: + \: n) - n \\<br /> = \frac{(n+1)n}{2} - n \\<br /> = \frac{n^2-n}{2}$$

 

[mathman]

What is the point? Standard combinatorial formula $$\binom{n}{k}=\frac{n!}{k!(n-k)!}$$.

 

[archaic]

What is the point? Standard combinatorial formula $$\binom{n}{k}=\frac{n!}{k!(n-k)!}$$.

well, call it a high schooler's curiosity and then impulse