I’ve been asked to find the coefficient of x^16y^4 in the expansion of (3x-1/9y)^20. I have the answer, i just don't understand completely.?

Question

How many 4 letter combinations can be selected from 20 letters….using the standard formula for calculating combinations this is 4845.

Now I’ve been asked to find the coefficient of x^16y^4 in the expansion of (3x-1/9y)^20. This is a past exam paper so I have the answer but I do not understand the part where we get a 3^-8.

4845*3^16*3^-8=4845*3^8=31788045

Where does the -1/9^4 go and why is there a 3^-8?
Can you please explain?

Thank you in advance
Donna

Answer 1

Expansion is F = f(x,y) = [ 3x + (-1/9)y]^20. Now (a + b)^n = sum, k = 0,1,2,...,n. of [nCk][a^(n-k)][b^k]. Here, we

have a =3x, b =(-1/9)y, n =20 and are asked for the coefficient of (x^16)(y^4). It will be [(20)C(4)][3^(20-4)][-(1/9)^4]

= {(20)!/[(20-4)!(4)!]}{[3^16][(1/9)^4]} = {20*19*18*17/[4*3*2]}[9^4} = 15*17*19(9^4) = 4845*6561 = 31,788,045.

Answer 2

Well,



the expansion can be written :



(3x - 1/9y)^20 = Σ (f from 0 to 20) C(20,k) (3x)^k * (-(1/9)y)^(20-k)



so the factors of the coeff. in x^16y^(20-16) are made of:

C(20,16) = C(20,4) = 20*19*18*17/(4*3*2*1) = 4845

3^16 = 43046721

(-1/9)^4 = 6561

therefore the coeff is :

C = 31,788,045



et voilà !!



hope it' ll help !!

Answer 3

We have (a-b)^20

a = 3x

b = 1/(9y)





Coefficient of (3x)^16 is 20C4 (3x)^16 (1/9^4) y^4

20C4 = 20x19x18x17 / 4x3x2x1 = 4,845

Coefficient of (x)^16 y^4 is (4845)(3)^16 / 9^4 = (4845) 3^16 / 3^8

= (4845) 3^8

= 31,788,045