The coefficient of $$x^4$$ in the expansion of $$(1 + x + x^2 + x^3)^6$$ in powers of $$x$$, is.......

Let us first simplify the expression inside the parentheses by factoring it. We notice that:

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

Therefore, we can rewrite the entire expression raised to the power of 6 as:

$$(1 + x + x^2 + x^3)^6 = \left[(1 + x)(1 + x^2)\right]^6 = (1 + x)^6(1 + x^2)^6$$

Now, we expand each of these two components using the Binomial Theorem:

$$(1 + x)^6 = \binom{6}{0} + \binom{6}{1}x + \binom{6}{2}x^2 + \binom{6}{3}x^3 + \binom{6}{4}x^4 + \dots$$

$$(1 + x^2)^6 = \binom{6}{0} + \binom{6}{1}x^2 + \binom{6}{2}x^4 + \dots$$

Let us evaluate the specific combinations (binomial coefficients) needed for our calculations:

$$\binom{6}{0} = 1$$

$$\binom{6}{1} = 6$$

$$\binom{6}{2} = \frac{6 \times 5}{2} = 15$$

$$\binom{6}{3} = \frac{6 \times 5 \times 4}{6} = 20$$

$$\binom{6}{4} = \binom{6}{2} = 15$$

Substituting these values back into our expansions yields:

$$(1 + x)^6 = 1 + 6x + 15x^2 + 20x^3 + 15x^4 + \dots$$

$$(1 + x^2)^6 = 1 + 6x^2 + 15x^4 + \dots$$

We want to find the coefficient of $$x^4$$ in the product of these two series:

$$\left(1 + 6x + 15x^2 + 20x^3 + 15x^4 + \dots\right) \times \left(1 + 6x^2 + 15x^4 + \dots\right)$$

Let us collect all combinations of terms from the first and second parentheses whose powers of $$x$$ add up to exactly 4:

1) The $$x^4$$ term from the first parentheses multiplied by the constant term ($$1$$) from the second:

$$15x^4 \times 1 = 15x^4$$

2) The $$x^2$$ term from the first parentheses multiplied by the $$x^2$$ term from the second:

$$15x^2 \times 6x^2 = 90x^4$$

3) The constant term ($$1$$) from the first parentheses multiplied by the $$x^4$$ term from the second:

$$1 \times 15x^4 = 15x^4$$

Now, we sum the coefficients of these three cases together to get the total coefficient of $$x^4$$:

$$\text{Coefficient of } x^4 = 15 + 90 + 15 = 120$$