Find the coefficient of x^{5} y¹0} in the expansion of (3 x+2 y)¹5}

Correct Answer is : 15C1035210{ }¹5} {C}₁0} 3^{5} 2¹0}15C1035210

Correct Answer is : 15C1035210{ }¹5} {C}₁0} 3^{5} 2¹0}15C1035210

Test Index

UPSC NDA Math Model Paper 10

Question : 83 of 120

Marks: +1, -0

x^{5} y^{10}

(3 x+2 y)^{15}

Solution:

For the given expression

(3 x+2 y)^{15}, n=15

({r}+1)

th term

{T}_{{r}+1}={ }^{15} {C}_{{r}}(3 {x})^{15-{r}}(2 {y})^{{r}}

For the coefficient of

{x}^{5} {y}^{10}, {r}

must be 10

{T}_{10+1}={ }^{15} {C}_{10}(3 {x})^{15-10}(2 {y})^{10}

{T}_{10+1}={ }^{15} {C}_{10}(3 {x})^{5}(2 {y})^{10}

{T}_{10+1}={ }^{15} {C}_{10} 3^{5} 2^{10} {x}^{5} {y}^{10}

So the coefficient of

{x}^{5} {y}^{10}

is equal to the

{ }^{15} {C}_{10} 3^{5} 2^{10}

Go to Question:

Prev Question

Next Question