Question

Find the coefficient of a^5b^7 in the expansion of (a + b)^12.

Answer 1

so general formula for comuting (x+y)^n is :
\[(x+y)^{n}=\sum_{j=0}^{n}\left(\begin{matrix}n \\ j\end{matrix}\right)x^{n-j}y^{j}\]
so you need the one in front of a^5b^7
so that would be substituting for n=12,j=7,x=a , y=b

Answer 2

We werent really taught that formula so =/ ...is there another way of going about it?

Answer 3

hm how do you solve it then ?

Answer 4

what formulas u have

Answer 5

I've never come across such a question hence putting it on here...but I usually use the simple finding of the factorials then later on multiplying whilst finding the value of a certain coefficient

Answer 6

ok wait when u look at that formula seems complicated thats just general formula for expansion lets find the coeficient

Answer 7

Okay, but like since there are no values for a and b I can't do it my way. That formula helps you find it ?

Answer 8

yeah we should be able to find it so that formula gives us every term in the expansion we want just one so we take just whats inside the sumation
so we would have :
\[\left(\begin{matrix}n \\ j\end{matrix}\right)x^{n-j}y^{j}\]

Answer 9

so pluging in our values we would have
\[\left(\begin{matrix}12 \\ 7\end{matrix}\right)a^{5}b^{7}\]

Answer 10

so what we left with is only the combinations
\[\left(\begin{matrix}12 \\ 7\end{matrix}\right)\] calculating that we will have the coeficient in front of that term

Answer 11

do you know to calculate that ?

Answer 12

Ohhh yeah I can do that. Thank youu

Answer 13

you are welcomed :)

Answer 14

But wait

Answer 15

Ohh nvm 792 got it

Answer 16

yeap thats correct :)