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7.Binomial Theorem
hard
The coefficient of $x ^7$ in $\left(1-x+2 x^3\right)^{10}$ is $........$.
A
$960$
B
$950$
C
$940$
D
$960$
(JEE MAIN-2023)
Solution
$\text { General term }=\frac{10 !}{r_{1} ! \cdot r_{2} ! \cdot r_{3} !}(-1)^{r_2} \cdot(2)^{r_3} x^{r_2+3 r_3}$
where $r_1+r_2+r_3=10$ and $r_2+3 r_3=7$
$\begin{array}{lll}r_1 & r_2 & r_3 \\ 3 & 7 & 0 \\ 5 & 4 & 1 \\ 7 & 1 & 2\end{array}$
Required coefficient
$=\frac{10 !}{3 ! .7 !}(-1)^7+\frac{10 !}{5 ! .4 !}(-1)^4(2)+\frac{10 !}{7 ! \cdot 2 !}(-1)^1(2)^2$
$=-120+2520-1440=960$
Standard 11
Mathematics
