Find the coefficient of ${x}^{20 }$ in ${({x}^{2}+{x}^{3}+{x}^{4}+{x}^{5}+{x}^{6})}^{5}$

Question

I saw a question in my textbook, the solution of this question exists in my textbook.

However , its solution is very long.I tried to solve it in different way but i do not know whether it is true or not.

The question: Find the coefficient of ${x}^{20 }$ in ${({x}^{2}+{x}^{3}+{x}^{4}+{x}^{5}+{x}^{6})}^{5}$

My solution:

Step 1-) ${({x}^{2}+{x}^{3}+{x}^{4}+{x}^{5}+{x}^{6})}^{5}$ is equal to ${[{x}^{2}(1+{x}+{x}^{2}+{x}^{3}+{x}^{4})]}^{5}$

Step 2-) ${[{x}^{2}(1+{x}+{x}^{2}+{x}^{3}+{x}^{4})]}^{5}$ is equal to ${x}^{10}$ ${(\frac{1- {x}^{5}}{1-x})}^{5}$

Step 3-) We should ${x}^{10}$ in the expansion of ${(\frac{1- {x}^{5}}{1-x})}^{5}$. We know that ${(\frac{1- {x}^{5}}{1-x})}^{5}$ is equal to ${{(1- {x}^{5})^{5}}({1-x})}^{-5}$ .So, we should find the coefficient of ${x}^{2} $ in the expansion of $({1-x})^{-5}$

Step 4-) The formula for finding the coefficient is $C(5+2-1,2)=15$

Is my solution correct ? If it is not , can you give me hints ,shortcuts or full solution. Thanks for your helps..

Answer 1

I'm afraid @lulu is right. Let $[x^n]f(x)$ denote the $x^n$ coefficient in $f(x)$. As you've noted, you want to calculate $[x^{10}](1-x^5)^5(1-x)^{-5}$. This is not, however, equal to $[x^2](1-x)^{-5}$. Instead it's$$\begin{align}&[x^{10}](1-x^5)^5[x^0](1-x)^{-5}+[x^5](1-x^5)^5[x^5](1-x)^{-5}+[x^0](1-x^5)^5[x^{10}](1-x)^{-5}\\&=\binom52\cdot1-5\cdot\binom94+1\cdot\binom{14}{4}\\&=10\cdot1-5\cdot126+1\cdot1001\\&=381.\end{align}$$

Answer 2

Hint. Once you have reached the second step, i.e. $P(x)=\left(1+x+x^2+x^3+x^4\right)^5\cdot x^{10}$, let me introduce the following problem: You are given the numbers $0,1,2,3,4$ and have to add five of them - repetition is allowed - in order to obtain $10$. How many options do you have?

Advice: Be careful when counting :)