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Question: If m = \(nC_{2}\), then \(mC_{2}\) is equal to...

If m = nC2nC_{2}, then mC2mC_{2} is equal to

A

n+1C4n + 1C_{4}

B

3xnC43x^{n}C_{4}

C

3xn+1C43x^{n + 1}C_{4}

D

None

Answer

3xn+1C43x^{n + 1}C_{4}

Explanation

Solution

Here,

m = nC2=n(n−1)2nC_{2} = \frac{n(n - 1)}{2} ... (1)

∴ mC2=m(m−1)2mC_{2} = \frac{m(m - 1)}{2}

= n(n−1)2 6mu{n(n−1)2 6mu− 6mu1}2\frac{\frac{n(n - 1)}{2}\mspace{6mu}\left\{ \frac{n(n - 1)}{2}\mspace{6mu} - \mspace{6mu} 1 \right\}}{2}

= n(n−1)(n2−n−2)8\frac{n(n - 1)(n^{2} - n - 2)}{8}

= n(n−1)(n−2)(n+1)8\frac{n(n - 1)(n - 2)(n + 1)}{8}

= (n+1)n(n−1)(n−2)24x3\frac{(n + 1)n(n - 1)(n - 2)}{24}x3

= 3(n+1)n(n−1)(n−2)∠4\frac{(n + 1)n(n - 1)(n - 2)}{\angle 4}

= 3 x n+1C4n + 1C_{4}