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B
2m+1C2
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C
3m+1C4
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D
3m+1C2
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Solution
The correct option is C3m+1C4 We know that, n=mC2=m(m−1)2 Now, nC2=n(n−1)2⇒nC2=m(m−1)(m(m−1)2−1)2×2⇒nC2=m(m−1)(m2−m−2)8⇒nC2=m(m−1)(m+1)(m−2)8⇒nC2=4!×(m+1)m(m−1)(m−2)(m−3)!8×4!(m−3)!⇒nC2=4!×m+1C48∴nC2=3m+1C4