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# Polynomials

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What is the coefficient of x^3 in (x^4 + x^3 + x^2 + x + 1)^5?

May 29, 2021

### 2+0 Answers

#1
+26321
+1

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 $$(x^4 + x^3 + x^2 + x + 1)^5$$?

$$(x^4 + x^3 + x^2 + x + 1)^5 =(1 + x^2 + x^3 + x^4)^5$$

$$\begin{array}{l} \mathbf{(1 + x^2 + x^3 + x^4)^2} \\ =(1 + x^2 + x^3 + x^4)(1 + x^2 + x^3 + x^4) \\ \end{array}\\ \begin{array}{|lcccccccccc|} \hline &=& &1&+&x &+&x^2 &+&x^3+\dots \\ && & &+&x &+&x^2 &+&x^3+\dots \\ && & & & &+&x^2 &+&x^3+\dots \\ && & & & && &+&x^3+\dots \\ \hline &=& &1&+&2x&+&3x^2&+&4x^3+\dots \\ \hline \end{array}$$

$$\begin{array}{l} \mathbf{(1 + x^2 + x^3 + x^4)^3} \\ =(1 + 2x+3x^2+4x^3 +\dots)(1 + x^2 + x^3 + x^4) \\ \end{array}\\ \begin{array}{|lcccccccccc|} \hline &=& &1&+&x &+&x^2 &+&x^3+\dots \\ && & &+&2x &+&2x^2 &+&2x^3+\dots \\ && & & & &+&3x^2 &+&3x^3+\dots \\ && & & & && &+&4x^3+\dots \\ \hline &=& &1&+&3x&+&6x^2&+&10x^3+\dots \\ \hline \end{array}$$

$$\begin{array}{l} \mathbf{(1 + x^2 + x^3 + x^4)^4} \\ =(1 + 3x+6x^2+10x^3 +\dots)(1 + x^2 + x^3 + x^4) \\ \end{array}\\ \begin{array}{|lcccccccccc|} \hline &=& &1&+&x &+&x^2 &+&x^3+\dots \\ && & &+&3x &+&3x^2 &+&3x^3+\dots \\ && & & & &+&6x^2 &+&6x^3+\dots \\ && & & & && &+&10x^3+\dots \\ \hline &=& &1&+&4x&+&10x^2&+&20x^3+\dots \\ \hline \end{array}$$

$$\begin{array}{l} \mathbf{(1 + x^2 + x^3 + x^4)^5} \\ =(1 + 4x+10x^2+20x^3 +\dots)(1 + x^2 + x^3 + x^4) \\ \end{array}\\ \begin{array}{|lcccccccccc|} \hline &=& &1&+&x &+&x^2 &+&x^3+\dots \\ && & &+&4x &+&4x^2 &+&4x^3+\dots \\ && & & & &+&10x^2 &+&10x^3+\dots \\ && & & & && &+&20x^3+\dots \\ \hline &=& &1&+&{\color{red}5}x&+&15x^2&+&{\color{red}35}x^3+\dots \\ \hline \end{array}$$

May 30, 2021
#2
+598
+1

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.

May 30, 2021