The coefficient of $$x^{1012}$$ in the expansion of $$(1 + x^n + x^{253})^{10}$$, (where $$n \leq 22$$ is any positive integer), is:

The given expression is $$(1 + x^n + x^{253})^{10}$$, and we need to find the coefficient of $$x^{1012}$$ where $$n$$ is a positive integer with $$n \leq 22$$.

In the expansion of $$(1 + x^n + x^{253})^{10}$$, each term arises from selecting one of the three options—$$1$$, $$x^n$$, or $$x^{253}$$—from each of the 10 factors. The general term is given by the multinomial coefficient:

$$\text{term} = \binom{10}{a, b, c} \cdot (1)^a \cdot (x^n)^b \cdot (x^{253})^c = \binom{10}{a, b, c} \cdot x^{n b + 253 c}$$

where $$a$$, $$b$$, and $$c$$ are non-negative integers satisfying $$a + b + c = 10$$, and $$\binom{10}{a, b, c} = \frac{10!}{a! b! c!}$$.

We require the exponent of $$x$$ to be 1012:

$$n b + 253 c = 1012$$

and the constraint $$a + b + c = 10$$ implies $$b + c \leq 10$$ since $$a = 10 - b - c \geq 0$$.

Given $$n \leq 22$$, we solve for non-negative integers $$b$$ and $$c$$ such that $$n b + 253 c = 1012$$ and $$b + c \leq 10$$. Rearranging for $$b$$:

$$b = \frac{1012 - 253 c}{n}$$

Since $$b$$ must be a non-negative integer, $$1012 - 253 c$$ must be divisible by $$n$$ and non-negative. Also, $$253 c \leq 1012$$ implies $$c \leq \frac{1012}{253} \approx 4.00$$, so $$c$$ can be 0, 1, 2, 3, or 4.

We check each possible $$c$$:

• For $$c = 0$$: $$b = \frac{1012}{n}$$. Then $$b + c = b \leq 10$$ implies $$\frac{1012}{n} \leq 10$$, so $$n \geq 101.2$$. But $$n \leq 22$$, contradiction. No solution.
• For $$c = 1$$: $$b = \frac{759}{n}$$. Then $$b + c = b + 1 \leq 10$$ implies $$b \leq 9$$, so $$\frac{759}{n} \leq 9$$ gives $$n \geq 84.333$$. Factorizing 759 = 3 × 11 × 23, divisors ≤22 are 1, 3, 11. But $$n \geq 84.333 > 22$$, no divisor satisfies. No solution.
• For $$c = 2$$: $$b = \frac{506}{n}$$. Then $$b + c = b + 2 \leq 10$$ implies $$b \leq 8$$, so $$\frac{506}{n} \leq 8$$ gives $$n \geq 63.25$$. Factorizing 506 = 2 × 11 × 23, divisors ≤22 are 1, 2, 11, 22. But $$n \geq 63.25 > 22$$, no divisor satisfies. No solution.
• For $$c = 3$$: $$b = \frac{253}{n}$$. Then $$b + c = b + 3 \leq 10$$ implies $$b \leq 7$$, so $$\frac{253}{n} \leq 7$$ gives $$n \geq 36.142$$. Factorizing 253 = 11 × 23, divisors ≤22 are 1, 11. But $$n \geq 36.142 > 22$$, no divisor satisfies. No solution.
• For $$c = 4$$: $$b = \frac{0}{n} = 0$$. Then $$b + c = 0 + 4 = 4 \leq 10$$, and $$a = 10 - 0 - 4 = 6$$. The exponent is $$n \cdot 0 + 253 \cdot 4 = 1012$$, which matches. This works for any $$n$$.

No other $$c$$ yields a solution given $$n \leq 22$$. Thus, the only solution is $$a = 6$$, $$b = 0$$, $$c = 4$$.

The coefficient is the multinomial coefficient:

$$\binom{10}{6, 0, 4} = \frac{10!}{6! 0! 4!}$$

Since $$0! = 1$$, this simplifies to:

$$\frac{10!}{6! 4!} = \binom{10}{4}$$

as $$\binom{10}{4} = \frac{10!}{4! 6!}$$.

Comparing with the options:

• A. $$^{253}C_4$$
• B. $$^{10}C_4$$
• C. $$4n$$
• D. $$1$$

The coefficient $$\binom{10}{4}$$ corresponds to option B.

Hence, the correct answer is Option B.