Combinatorics-binomial expansion?

[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!