In the binomial expansion of \((1+x)^{15}\), the coefficients of \(\mathrm{x}^{\mathrm{r}}\) and \(\mathrm{x}^{\mathrm{r}+3}\) are equal. Then, \(r\) is

\(\begin{array}{lc}\text { Given, } & (1+\mathrm{x})^{15} \\ \text { Now, } & \mathrm{T}_{\mathrm{r}+1}={ }^{15} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}} \\ \text { and } & \mathrm{T}_{(\mathrm{r}+3)+1}={ }^{15} \mathrm{C}_{\mathrm{r}+3} \mathrm{x}^{\mathrm{r}+3}\end{array}\)
According to question
coefficient of \(\mathrm{x}^{\mathrm{r}}=\) coefficient of \(\mathrm{x}^{\mathrm{r}+3}\)
\[
\begin{aligned}
&\Rightarrow \quad{ }^{15} C_{r}={ }^{15} C_{r+3} \\
&\Rightarrow \quad \frac{15 !}{r !(15-r) !}=\frac{15 !}{(r+3) !(12-r) !} \\
&\Rightarrow \frac{1}{(15-r)(14-r)(13-r)} \\
&\Rightarrow(r+1)(r+2)(r+3)=\frac{1}{(r+3)(r+2)(r+1)} \\
&\Rightarrow\left(r^{2}+3 r+2\right)(r+3)=\left(210-29 r+r^{2}\right)(13-r)
\end{aligned}
\]
\(\begin{array}{lc}\Rightarrow & r^{3}+3 r^{2}+2 r+3 r^{2}+9 r+6 \\ & =2930-377 r+13 r^{2}-210 r+29 r^{2}-r^{3} \\ \Rightarrow & 2 r^{3}-36 r^{2}+598 r-2924=0 \\ \Rightarrow & r^{3}-18 r^{2}+299 r-1462=0 \\ \Rightarrow & (r-6)\left(r^{2}-12 r+227\right)=0 \\ \Rightarrow r & \text { and } r^{2}-12 r+227=0 \text { gives imaginary } \\ \text { roots. }\end{array}\)
Alternate Method
\({ }^{15} \mathrm{C}_{\mathrm{r}}={ }^{15} \mathrm{C}_{\mathrm{r}+3}\)
\(\Rightarrow \quad \mathrm{r}+(\mathrm{r}+3)=15 \quad\left(\because{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{x}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{y}}\right)\)
\(\Rightarrow \quad 2 \mathrm{r}+3=15\)
\[
\Rightarrow \quad \mathrm{x}+\mathrm{y}=\mathrm{n}
\]
\(\Rightarrow \quad 2 \mathrm{r}=12\)
\[
\Rightarrow \quad \mathrm{r}=6
\]