Find the coefficient of $x^{3}$ in the expansion of $(1 + x + x^{2})^{5}$

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Solution

$(1 + x + x^{2})^{5} = \sum_{r_{1} + r_{2} + r_{3} = 5} \frac{5!}{r_{1}! \times r_{2}! \times r_{3}!} \times (1)^{r_{1}} \times (x)^{r_{2}} (x^{2})^{r_{3}}$

We want $r_{2} + 2 r_{3} = 3$ (power of x=3)
The values $(r_{1}, r_{2}, r_{3})$ can take are (3,1,1) and (2,3,0)
So the coefficient is the sum of $\frac{5!}{3! \times 1! \times 1!} + \frac{5!}{2! \times 3! \times 0!} = 20 + 10 = 30$

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