Combinatorics: solving for coefficient of x^n term

AI Thread Summary
To find the coefficient of the x^10 term in the equation f(x)=(x+x^2+x^3+x^4+x^5+x^6)^3, the correct approach involves using the binomial theorem. The function can be rewritten as f(x) = x^3 * ((1-x^6)/(1-x))^3. Initially, there was confusion regarding the application of the binomial theorem, but clarification helped resolve the issue. The expected coefficient for the x^10 term is 27, which aligns with a previous problem. Understanding the binomial expansion is crucial for solving this combinatorial problem effectively.
Armbru35
Messages
7
Reaction score
0
Hi, I'm currently taking a Discrete Mathematics class and cannot seem to work out this one problem, we need to find the x^10 term in order to determine its coefficient of the equation f(x)=(x+x^2+x^3+x^4+x^5+x^6)^3 I know the answer is to be 27 from a previous problem (we are to use this method to verify our answer) but I can't seem to figure it out. I started with thinking of trying to solve x^3/((1-x)^3)-the sum of x^n starting with n≥7, but that doesn't seem to be working. Any suggestions would be appreciated!
 
Physics news on Phys.org
Hint:

f(x) = x^3 \left( \frac{1-x^6}{1-x} \right) ^ 3
 
Ahhh...I was confused for a second but I was doing the binomial theorem wrong. Thank you so much that helped tremendously!
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
Back
Top