Q56 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics

${ }^{34} C_{10}+3 \cdot\left({ }^{34} C_9\right)+3 \cdot\left({ }^{34} C_g\right)+{ }^{34} C_7=$

  • A
    ${ }^{39} \mathrm{C}_{10}$
  • B
    ${ }^{36} \mathrm{C}_{10}$
  • Correct answer
    C
    ${ }^{37} \mathrm{C}_{10}$
  • D
    ${ }^{35} \mathrm{C}_{10}$

Solution

$\begin{aligned}{ }^{34} C_{10}+ & 3\left({ }^{34} C_9\right)+3\left({ }^{34} C_8\right)+{ }^{34} C_7 \\ = & \left({ }^{34} C_{10}+{ }^{34} C_9\right)+2 \cdot\left({ }^{34} C_9\right)+2 \cdot\left({ }^{34} C_8\right) \\ & +\left({ }^{34} C_8+{ }^{34} C_7\right) \\ = & { }^{35} C_{10}+2 \cdot\left({ }^{35} C_9\right)+{ }^{35} C_8 \\ = & { }^{35} C_{10}+{ }^{35} C_9+{ }^{35} C_9+{ }^{35} C_8 \\ = & { }^{36} C_{10}+{ }^{36} C_9={ }^{37} C_{10}\end{aligned}$

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

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