If ${ }^{11} \mathrm{C}_4+{ }^{11} \mathrm{C}_5+{ }^{12} \mathrm{C}_6+{ }^{13} \mathrm{C}_7={ }^{14}…

If ${ }^{11} \mathrm{C}_4+{ }^{11} \mathrm{C}_5+{ }^{12} \mathrm{C}_6+{ }^{13} \mathrm{C}_7={ }^{14} \mathrm{C}_{\mathrm{r}}$, then value of $\mathrm{r}$ is
  1. 11
  2. 14
  3. 7
  4. 3

Solution

${ }^{11} \mathrm{C}_4+{ }^{11} \mathrm{C}_5+{ }^{12} \mathrm{C}_6+{ }^{13} \mathrm{C}_7={ }^{14} \mathrm{C}_{\mathrm{r}}$ We know that ${ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r$ $\begin{aligned} & \therefore{ }^{14} \mathrm{C}_{\mathrm{r}}=\left({ }^{11} \mathrm{C}_4+{ }^{11} \mathrm{C}_5\right)+{ }^{12} \mathrm{C}_6+{ }^{13} \mathrm{C}_7 \\ & =\left({ }^{12} \mathrm{C}_5+{ }^{12} \mathrm{C}_6\right)+{ }^{13} \mathrm{C}_7 \\ & ={ }^{13} \mathrm{C}_6+{ }^{13} \mathrm{C}_7={ }^{14} \mathrm{C}_7 \\ & \therefore \mathrm{r}=7 \end{aligned}$

Asked in: MHT CET 2021 (24 Sep Shift 1)

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