The coefficient of $x^{2} y^{3}$ in the expansion of $(1 - x + y)^{20}$ is

\frac{20!}{2! 3!}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- \frac{20!}{2! 3!}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{20!}{- 5! 2! 3!}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

none of these

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is B none of these
${(1 - x + y)}^{20} = \sum \frac{(20)!}{(r_{1})! (r_{2})! (r_{3})!} {1^{r} (- x)^{r_{2}} (y)^{r_{3}}}$
$r_{1} + r_{2} + r_{3} = 20$
To find the coefficient of $x^{2} y^{3}$
$\Rightarrow r_{2} = 2 : r_{3} = 3$
$&$ $r_{1} = 20 - (2 + 3) = 15$
$∴$ Coefficient $= \frac{20!}{(15!) (2!) (3!)}$
Hence, the answer is none of these.

Suggest Corrections

Similar questions

x^{20}

(1 + 3 x + 3 x^{2} + x^{3})^{20}

x^{2} y^{3}

(1 - x + y)^{20}

x^{2} y^{3}

(1 - x + y)^{20}

If rth term is the middle term in the expansion of ${(x^{2} - \frac{1}{2 x})}^{20}$ , then $(r + 3)^{t h}$ term is

x^{203}

(x - 1) (x^{2} - 2) (x^{3} - 3) \dots (x^{20} - 20)

Join BYJU'S Learning Program