Indian Institute of Foreign Trade 2009 Solved Paper

We need to find the coefficient of ${\mathit{x}}^7$ in the expansion $\left(1-{\mathit{x}}^2+{\mathit{x}}^3\right){\left(1+\mathit{x}\right)}^{10}$
Now,${\left(1+\mathit{x}\right)}^{10}$ will have all the powers of x from 0 to 10. Multiplying these powers by $1,{\mathit{x}}^2$ and ${\mathit{x}}^3$ will yield different results but we are interested in finding only the coefficient of ${\mathit{x}}^7$ When we multiply ${\mathit{x}}^7$ of ${\left(1+\mathit{x}\right)}^{10}$ by $1,{\mathit{x}}^5$ of ${\left(1+\mathit{x}\right)}^{10}$ by ${\mathit{x}}^2$ and ${\mathit{x}}^4$ of ${\left(1+\mathit{x}\right)}^{10}$ by ${\mathit{x}}^3$.we will get ${\mathit{x}}^7$
coefficient of ${\mathit{x}}^7$ in ${\left(1+\mathit{x}\right)}^{10}$ is $10{\mathit{C}}_7=120$,
coefficient of ${\mathit{x}}^5$ in ${\left(1+\mathit{x}\right)}^{10}$ is $10{\mathit{C}}_5=252$,
coefficient of ${\mathit{x}}^4$ in ${\left(1+\mathit{x}\right)}^{10}$ is $10{\mathit{C}}_4=210$
adding 120 and 210 and subtracting(since ${\mathit{x}}^2$ has a negative sign) 252 we get coefficient of ${\mathit{x}}^7$ as 78
Therefore our answer is option 'B'