Haryana Police Constable Model Paper 8

Concept:
The general term of the expansion of ${\left(\mathrm{x}+\mathrm{a}\right)}^{\mathrm{n}}$ is usually denoted by ${\mathrm{T}}_{\mathrm{r}+1}={}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{\mathrm{x}}^{\mathrm{n}-\mathrm{r}}{\mathrm{a}}^{\mathrm{r}}$

Calculation:
For the given expression ${\left(3\mathrm{x}+2\mathrm{y}\right)}^{15},\mathrm{n}=15$
${\left(\mathrm{r}+1\right)}^{\mathrm{th}}$ term
${\mathrm{T}}_{\mathrm{r}+1}={}^{15}{\mathrm{C}}_{\mathrm{r}}{\left(3\mathrm{x}\right)}^{15-\mathrm{r}}{\left(2\mathrm{y}\right)}^{\mathrm{r}}$
For the coefficient of ${\mathrm{x}}^5{\mathrm{y}}^{10},\mathrm{r}$ must be $10$ .
${\mathrm{T}}_{10+1}={}^{15}{\mathrm{C}}_{10}{\left(3\mathrm{x}\right)}^{15-10}{\left(2\mathrm{y}\right)}^{10}$
${\mathrm{T}}_{10+1}={}^{15}{\mathrm{C}}_{10}{\left(3\mathit{x}\right)}^5{\left(2\mathrm{y}\right)}^{10}$
${\mathrm{T}}_{10+1}={}^{15}{\mathrm{C}}_{10}{3}^5{2}^{10}{\mathrm{x}}^5{\mathrm{y}}^{10}$
So the coefficient of ${\mathrm{x}}^5{\mathrm{y}}^{10}$ is equal to the ${}^{15}{\mathrm{C}}_{10}{3}^5{2}^{10}$