What is the coefficient of x3 y4 in (2 x+3 y2)5 ?

Question

What is the coefficient of x3 y4 in (2 x+3 y2)5 ? (A) 240 (B) 360 (C) 720 (D) 1080

Tardigrade Admin · Accepted Answer

T r = n C r -1(2 x ) r -1(3 y 2) n - r +1 T 4= T 3+1= 5 C 3(2 x )3(3 y 2)2 =(5 !/3 ! 2 !) 23 . x3 . 9 y4=(5.4/2.1) × 8 × 9 × x3 y4=720 x3 y4 ∴ Coefficient of x3 y4=720

Q. What is the coefficient of $x^{3} y^{4}$ in $(2 x + 3 y^{2})^{5} ?$

240

360

720

1080

$T_{r} =^{n} C_{r - 1} (2 x)^{r - 1} (3 y^{2})^{n - r + 1}$
$T_{4} = T_{3 + 1} =^{5} C_{3} (2 x)^{3} (3 y^{2})^{2}$
$= \frac{5 !}{3 ! 2 !} 2^{3} . x^{3} .9 y^{4} = \frac{5.4}{2.1} \times 8 \times 9 \times x^{3} y^{4} = 720 x^{3} y^{4}$
$∴$ Coefficient of $x^{3} y^{4} = 720$

Q. What is the coefficient of x3y4 in (2x+3y2)5?

240

360

720

1080

Tr​=nCr−1​(2x)r−1(3y2)n−r+1 T4​=T3+1​=5C3​(2x)3(3y2)2 =3!2!5!​23.x3.9y4=2.15.4​×8×9×x3y4=720x3y4 ∴ Coefficient of x3y4=720

Q. What is the coefficient of $x^{3} y^{4}$ in $(2 x + 3 y^{2})^{5} ?$

$T_{r} =^{n} C_{r - 1} (2 x)^{r - 1} (3 y^{2})^{n - r + 1}$
$T_{4} = T_{3 + 1} =^{5} C_{3} (2 x)^{3} (3 y^{2})^{2}$
$= \frac{5 !}{3 ! 2 !} 2^{3} . x^{3} .9 y^{4} = \frac{5.4}{2.1} \times 8 \times 9 \times x^{3} y^{4} = 720 x^{3} y^{4}$
$∴$ Coefficient of $x^{3} y^{4} = 720$