+26393 What is the coefficient of \(\color{red}x\) in \((x^4 + x^3 + x^2 + x + 1)^5\)?
What is the coefficient of \(\color{red}x^3\) in \((x^4 + x^3 + x^2 + x + 1)^5\)?
\((x^4 + x^3 + x^2 + x + 1)^5 =(1 + x^2 + x^3 + x^4)^5\)
\(\begin{array}{l} \mathbf{(1 + x^2 + x^3 + x^4)^2} \\ =(1 + x^2 + x^3 + x^4)(1 + x^2 + x^3 + x^4) \\ \end{array}\\ \begin{array}{|lcccccccccc|} \hline &=& &1&+&x &+&x^2 &+&x^3+\dots \\ && & &+&x &+&x^2 &+&x^3+\dots \\ && & & & &+&x^2 &+&x^3+\dots \\ && & & & && &+&x^3+\dots \\ \hline &=& &1&+&2x&+&3x^2&+&4x^3+\dots \\ \hline \end{array} \)
\(\begin{array}{l} \mathbf{(1 + x^2 + x^3 + x^4)^3} \\ =(1 + 2x+3x^2+4x^3 +\dots)(1 + x^2 + x^3 + x^4) \\ \end{array}\\ \begin{array}{|lcccccccccc|} \hline &=& &1&+&x &+&x^2 &+&x^3+\dots \\ && & &+&2x &+&2x^2 &+&2x^3+\dots \\ && & & & &+&3x^2 &+&3x^3+\dots \\ && & & & && &+&4x^3+\dots \\ \hline &=& &1&+&3x&+&6x^2&+&10x^3+\dots \\ \hline \end{array}\)
\(\begin{array}{l} \mathbf{(1 + x^2 + x^3 + x^4)^4} \\ =(1 + 3x+6x^2+10x^3 +\dots)(1 + x^2 + x^3 + x^4) \\ \end{array}\\ \begin{array}{|lcccccccccc|} \hline &=& &1&+&x &+&x^2 &+&x^3+\dots \\ && & &+&3x &+&3x^2 &+&3x^3+\dots \\ && & & & &+&6x^2 &+&6x^3+\dots \\ && & & & && &+&10x^3+\dots \\ \hline &=& &1&+&4x&+&10x^2&+&20x^3+\dots \\ \hline \end{array}\)
\(\begin{array}{l} \mathbf{(1 + x^2 + x^3 + x^4)^5} \\ =(1 + 4x+10x^2+20x^3 +\dots)(1 + x^2 + x^3 + x^4) \\ \end{array}\\ \begin{array}{|lcccccccccc|} \hline &=& &1&+&x &+&x^2 &+&x^3+\dots \\ && & &+&4x &+&4x^2 &+&4x^3+\dots \\ && & & & &+&10x^2 &+&10x^3+\dots \\ && & & & && &+&20x^3+\dots \\ \hline &=& &1&+&{\color{red}5}x&+&15x^2&+&{\color{red}35}x^3+\dots \\ \hline \end{array}\)
+605 Consider the following scenario: You have 5 baskets. At each basket you can pick anywhere from 0 to 4 apples. How many ways can you pick 3 apples? The function given is the generating function for this question, and so it suffices just to solve the scenario.
$(0,0,1,1,1), (0,0,0,1,2), (0,0,0,0,3)$ are the only ones that work. So the answer is $\frac{5!}{2!3!}+\frac{5!}{3!1!1!}+\frac{5!}{4!1!}=10+20+5=\boxed{35}$.
Alternatively:
See if you can find a pattern in @heureka's answer -- prove it.
