If ${ }^9 C_3+{ }^9 C_5={ }^{10} C_r$ for some $r \in \mathbb{N}$, then $r=$
If ${ }^9 C_3+{ }^9 C_5={ }^{10} C_r$ for some $r \in \mathbb{N}$, then $r=$
- $3$
- $4$
- $5$
- $7$
Solution
${ }^n \mathrm{C}_r={ }^n \mathrm{C}_{n-r}$
$\begin{aligned} & { }^9 \mathrm{C}_5={ }^9 \mathrm{C}_4 \\ & { }^9 \mathrm{C}_3+{ }^9 \mathrm{C}_5={ }^9 \mathrm{C}_3+{ }^9 \mathrm{C}_4 \\ & ={ }^{10} \mathrm{C}_4\left(\because{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}={ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}}\right)\end{aligned}$
$r=4$
Asked in: AP EAMCET 2022 (05 Jul Shift 2)
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