Mathematics › Hyperbola › Series involving Product of Coefficients
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
11 14 7 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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