High School Combinations of n elements in pairs

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The discussion focuses on the combinations of n elements taken in pairs, represented as C^2_n. It establishes a recursive relationship showing that C^2_n can be derived from previous values, specifically C^2_{n-1} plus (n-1). The calculations demonstrate that C^2_5 equals 10, confirming the formula C^2_n = (1 + 2 + ... + (n-1)). The thread emphasizes the standard combinatorial formula for combinations, which is binomial coefficient notation. Overall, the conversation highlights the mathematical principles behind calculating combinations in pairs.
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<br /> 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&#039;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}<br />
 
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What is the point? Standard combinatorial formula \binom{n}{k}=\frac{n!}{k!(n-k)!}.
 
mathman said:
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
 
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