Q164 / 270 · NDA 2022 (Phase 1) · Mathematics

What is the value of $C(51,21)-C(51,22)+C(51,23)-$ $C(51,24)+C(51,25)-C(51,26)+C(51,27)-C(51,28)+$ $C(51,29)-C(51,30)$ ?

  • A
    $C(51,25)$
  • B
    $C(51,27)$
  • Correct answer
    C
    $C(51,51)-C(51,0)$
  • D
    $C(51,25)-C(51,27)$

Solution

$\begin{aligned} & { }^{51} C_{21}-{ }^{51} C_{22}+{ }^{51} C_{23}-{ }^{51} C_{24}+{ }^{51} C_{25}-{ }^{51} C_{26}+{ }^{51} C_{27}-{ }^{51} C_{28} \\ & +{ }^{51} C_{29}-{ }^{51} C^{30} \\ & ={ }^{51} C_{51-21}-{ }^{51} C_{51-22}+{ }^{51} C_{51-23}-{ }^{51} C_{51-24}+{ }^{51} C_{51-25}- \\ & { }^{51} C_{26}+{ }^{51} C_{27}-{ }^{51} C_{28}+{ }^{51} C_{29}-{ }^{51} C_{30} \quad\left[\because{ }^{\mathrm{n}} C_{\mathrm{r}}={ }^{\mathrm{n}} C_{\mathrm{n}-1}\right] \\ & ={ }^{51} C_{30}-{ }^{51} C_{29}-{ }^{51} C_{28}-{ }^{51} C_{27}+{ }^{51} C_{26}-{ }^{51} C_{26}+{ }^{51} C_{27}-{ }^{51} C_{28} \\ & +{ }^{51} C_{29}-{ }^{51} C_{30} \\ & =0 \\ & \text {Now, }{ }^{51} C_{51}-{ }^{51} C_0=1-1=0\end{aligned}$

Hence, the correct answer is option: 'C'.

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