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Given nC1 = nC2.The formula for combinations is nCr = n! / (r! * (n-r)!).So, nC1 = n! / (1! * (n-1)!) = n.And nC2 = n! / (2! * (n-2)!) = (n * (n-1)) / 2.Setting them equal: n = (n * (n-1)) / 2.Since n must be a positive integer for nC1 and nC2 to be defined (n ≥ 2), we can divide by n:1 = (n-1) / 2.2 = n - 1.n = 3.Now, we check which option is true for n = 3:A: 2n = n => 2(3) = 3 => 6 = 3 (False)B: 2n = n(n + 1) => 2(3) = 3(3 + 1) => 6 = 3(4) => 6 = 12 (False)C: 2n = n(n - 1) => 2(3) = 3(3 - 1) => 6 = 3(2) => 6 = 6 (True)D: 2n = m(m - 1) is not directly comparable without a value for 'm', but if we interpret 'm' as 'n', it's the same as C.