Finding the number of possible solutions for a system of equations using the ball-dash method

Question

Determine the coefficient of $x^5$ in the expansion of $(x^2+2x+5)^5$.

By Leibniz's polynomial, we have $$\sum\frac{5!}{a!b!c!}x^{2a+b} 2^b 5^c.$$ Now we just need to analyze the possibilities for \begin{cases} 2a+b=5 \\ a+b+c=5\end{cases}
I would like to know how to know the number of possible solutions that this solution can have, in advance without trying to analyze each value that satisfies it. I thought about finding out using the ball-dash method, but I don't know how to do that...

Answer 1

I'm not familiar with the ball-dash method, but we can find the coefficient of $x^5$ for instance by applying the binomial theorem twice. We use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ of a series.

We obtain \begin{align*} \color{blue}{[x^5]}&\color{blue}{\left(x^2+2x+5\right)^5}\\ &=[x^5]\sum_{k=0}^{5}\binom{5}{k}\left(x^2+2x\right)^k5^{5-k}\\ &=\sum_{k=0}^{5}\binom{5}{k}[x^{5-k}](x+2)^k5^{5-k}\tag{1}\\ &=\sum_{k=0}^{5}\binom{5}{k}\binom{k}{5-k}2^{2k-5}5^{5-k}\tag{2}\\ &=\binom{5}{3}\binom{3}{2}2\cdot 5^2+\binom{5}{4}\binom{4}{1}2^3\cdot 5 +\binom{5}{5}\binom{5}{0}2^5\\ &=1\,500+800+32\\ &\,\,\color{blue}{=2\,332} \end{align*}

Comment:

• In (1) we apply the rule $[x^p]x^qA(x)=[x^{p-q}]A(x)$.

• In (2) we select the coefficient of $x^{5-k}$.