Combinatorics-binomial expansion?

[vela]

No, because, for example, the x² term in A(x) and the x³ term in B(x) will combine to contribute to the x⁵ term in A(x)B(x).

By the way, don't you see a contradiction in the answer you just gave, $126y^4$, and the formula you verified earlier, which is a sum of a bunch of terms?

 

[pupeye11]

Yeah, I forgot about the sum... But if I was working with just B(x) then my answer would be $$126y^4$$ but since I have A(x) I need to multiply together. Is there a theorem in combinatorics that makes this easy or do I just need to do it the long way?

 

[vela]

Well, there's the formula you verified earlier.

 

[pupeye11]

That is kind of the problem, I get that if I use the formula I'll have something that looks like this

$$\sum_{k>=0}(\sum_{i=0}^{k}a_{1}b_{4})x^5$$. How does this give me a number for a coefficient? Or do I take the coefficient at $$a_{1}$$ and $$b_{4}$$ and multiply them together. Then after doing that for all the combinations that create x^5 do I add them all together?

 

[vela]

How did you get this

$$\sum_{k>=0}(\sum_{i=0}^{k}a_{1}b_{4})x^5$$

from this

A(X)B(X) = $$\sum_{k>=0}(\sum_{i=0}a_{i}b_{k-i})x^k$$

The second sum sign in the answer should be from i=0 to k.

 

[pupeye11]

Yes, that came from the one you just copied and put above.

 

[vela]

I'm asking how you got the first from the second because they're inconsistent.

Expand this summation for k=5:

$$\sum_{i=0}^k a_i b_{k-i}$$

What do you get?

 

[pupeye11]

$$a_{0}b_{5}+a_{1}b_{4}+a_{2}b_{3}+a_{3}b_{2}+a_{4}b_{1}+a_{5}b_{0}$$

 

[vela]

Right, and that sum of six terms is the coefficient of the x⁵ term in A(x)B(x). In this case, all the a_i's are equal to 1. What are b_i's equal to?

 

[pupeye11]

The $$b_{i}'s$$ are going to be $$b_{5}=nCr(9,5), b_{4}=nCr(9,4),b_{3}=nCr(9,3),b_{2}=nCr(9,2),b_{1}=nCr(9,1),b_{0}=nCr(9,0)$$?

 

[vela]

Not quite. Remember, the coefficient of x^k includes everything that multiplies x^k when you expand (x+y)⁹.

 

[pupeye11]

Are you trying to say I need the corresponding y's in there too, like b5 would have a y^4 and b4 would have a y^5, so on so forth? Otherwise I am not following you on this one.

 

[vela]

Yes, exactly.

 

[pupeye11]

Alright so the answer will be $$nCr(9,5)y^4+nCr(9,4)y^5+nCr(9,3)y^6+nCr(9,2)y^7+nCr(9,1)y^8+nCr(9,0)y^9$$ right? and thank you for all your help!

 

[vela]

Yup, that's it. Good work!