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NDA 2026
Exam
NDA
Year
2026
Subject
Mathematics
Chapter
Permutation and Combination

NDA 2026 · Mathematics · Permutation and Combination

If n= ^ m C_2 then what is ^ n C_2 equal to ?

If n=mC2n={}^{m}C_2 then what is nC2{}^{n}C_2 equal to ?

Options

  1. Am+1C4{}^{m+1}C_4
  2. B2×m+1C42\times{}^{m+1}C_4
  3. C3×m+1C43\times{}^{m+1}C_4
  4. Dm+2C4{}^{m+2}C_4

Correct answer

C. 3×m+1C43\times{}^{m+1}C_4

Step-by-step solution

Given n=mC2=m(m−1)2n = ^{m}C_{2} = \dfrac{m(m-1)}{2}

Evaluating nC2=n(n−1)2^{n}C_{2} = \dfrac{n(n-1)}{2}

Substituting the value of nn:

nC2=m(m−1)2(m(m−1)2−1)2^{n}C_{2} = \dfrac{\dfrac{m(m-1)}{2} \left( \dfrac{m(m-1)}{2} - 1 \right)}{2}

nC2=m(m−1)(m2−m−2)8^{n}C_{2} = \dfrac{m(m-1)(m^2 - m - 2)}{8}

Factorizing m2−m−2=(m−2)(m+1)m^2 - m - 2 = (m-2)(m+1):

nC2=(m+1)m(m−1)(m−2)8^{n}C_{2} = \dfrac{(m+1)m(m-1)(m-2)}{8}

Multiplying and dividing by 33:

nC2=3×(m+1)m(m−1)(m−2)24^{n}C_{2} = 3 \times \dfrac{(m+1)m(m-1)(m-2)}{24}

nC2=3×m+1C4^{n}C_{2} = 3 \times ^{m+1}C_{4}

Answer: 3×m+1C43\times{}^{m+1}C_4
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