Discussion
Notation :
C(n, r) = n!/(r!*(n-r)!))
n! is factorial n.
|A| cardinality of set A
\(\overline{s}\) is negation of s
C(2n, 2) = .....................
Answer: C
The answer is 2C(n, 2) + \(n^{2}\).
\[ \begin{aligned} C(2n, 2) &= \frac{2n(2n-1)}{2} = 2n^{2} - n \\ 2C(n, 2) + n^{2} &= n(n-1) + n^{2} = 2n^{2} - n \end{aligned} \]
A combinatorial reading gives the same thing: split 2n objects into two groups of n. A pair is either inside the first group, inside the second (2C(n, 2) ways together) or has one object from each group (\(n \times n = n^{2}\) ways).
The others do not match: 2C(2n, 1) = 4n, nC(n, 1) + 2n = \(n^{2} + 2n\), and C(n, 2) + \(2n^{2}\) = \(2.5n^{2} - 0.5n\). Trying n = 2 confirms it: C(4, 2) = 6 and 2C(2, 2) + 4 = 6.
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