Write the correct alternative in the following:

Question:

Write the correct alternative in the following:

If $\frac{d}{d x}\left\{x^{n}-a_{1} x^{n-1}+a_{2} x^{n-2}+\ldots+(-1)^{n} a_{n}\right\}$

$e^{x}=x^{n} e^{x}$

Then the value of $a_{r}, 0

A. $\frac{\mathrm{n} !}{\mathrm{r} !}$

B. $\frac{(n-r) !}{r !}$

C. $\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) !}$

D. none of these

Solution:

Given:

$\frac{d}{d x}\left\{x^{n}-a_{1} x^{n-1}+a_{2} x^{n-2}+\cdots+(-1)^{n} a_{n}\right\} e^{x}=x^{n} e^{x}$

$\frac{d}{d x}\left\{a_{0}(-1)^{0} x^{n}+a_{1}(-1)^{1} x^{n-1}+a_{2}(-1)^{2} x^{n-2}+\cdots+(-1)^{n} a_{n}\right\} e^{x}$

$\frac{d}{d x}(x-1)^{n}$

$(x-1)^{n}=\sum_{k=0}^{n}\left(\begin{array}{l}n \\ k\end{array}\right) x^{n-k}(-1)^{k}$

So, at $k=r$;

$a_{r}=\left(\begin{array}{l}n \\ r\end{array}\right)$

Also, $\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right)=\left(\begin{array}{c}\mathrm{n} \\ \mathrm{n}-\mathrm{r}\end{array}\right)$

So, $a_{r}=\left(\begin{array}{c}n \\ n-r\end{array}\right)$

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