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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Permutation and Combination

NDA 2016 · Mathematics · Permutation and Combination

What is ^ 47 C_4+ ^ 51 C_3+∑_ j=2 ^5 ^ 52-j C_3 equal to?

What is 47C4+51C3+∑j=2552−jC3{ }^{47} \mathrm{C}_4+{ }^{51} \mathrm{C}_3+\sum_{\mathrm{j}=2}^5{ }^{52-\mathrm{j}} \mathrm{C}_3 equal to?

Options

  1. A52C4{ }^{52} \mathrm{C}_4
  2. B51C5{ }^{51} \mathrm{C}_5
  3. C53C4{ }^{53} \mathrm{C}_4
  4. D52C5{ }^{52} \mathrm{C}_5

Correct answer

A. 52C4{ }^{52} \mathrm{C}_4

Step-by-step solution

47C4+51C3+50C3+49C3+48C3+47C3=47C3+47C4+48C3+49C3+50C3+51C348C4+48C3+49C3+50C3+51C3(∵nCr+nCr+1=(n+1r+1))=49C4+49C3+50C3+51C3=50C4+50C3+51C3=51C4+51C3=52C4\begin{aligned} & { }^{47} C_4+{ }^{51} C_3+{ }^{50} C_3+{ }^{49} C_3+{ }^{48} C_3+{ }^{47} C_3 \\ & ={ }^{47} C_3+{ }^{47} C_4+{ }^{48} C_3+{ }^{49} C_3+{ }^{50} C_3+{ }^{51} C_3 \\ & { }^{48} C_4+{ }^{48} C_3+{ }^{49} C_3+{ }^{50} C_3+{ }^{51} C_3 \\ & \left(\because{ }^n C_r+{ }^n C_{r+1}=\binom{n+1}{r+1}\right) \\ & ={ }^{49} C_4+{ }^{49} C_3+{ }^{50} C_3+{ }^{51} C_3 \\ & ={ }^{50} C_4+{ }^{50} C_3+{ }^{51} C_3 \\ & ={ }^{51} C_4+{ }^{51} C_3 \\ & ={ }^{52} C_4 \end{aligned}
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