If the number of terms in the expansion of $$\left(1 - \frac{2}{x} + \frac{4}{y^2}\right)^n$$, $$x, y \neq 0$$, is 28, then the sum of the coefficients of all the terms in this expansion is

We begin with the multinomial expression $$\left(1-\dfrac{2}{x}+\dfrac{4}{y^2}\right)^n.$$

In every term of its expansion the exponents of the three individual parts $$1,\;-\dfrac{2}{x},\;\dfrac{4}{y^2}$$ must add up to the power $$n.$$

Let those exponents be $$r,\;s,\;t$$ respectively. We therefore have the condition $$r+s+t=n,$$ where $$r,\;s,\;t$$ are non-negative integers.

The total number of ordered triples $$(r,s,t)$$ satisfying $$r+s+t=n$$ equals the number of non-negative integral solutions of that equation. A standard result from combinatorics states that the number of such solutions is

$$\binom{n+3-1}{3-1}=\binom{n+2}{2}.$$

It is given that the expansion contains exactly 28 distinct terms, so

$$\binom{n+2}{2}=28.$$

Using the definition $$\displaystyle \binom{n+2}{2}=\dfrac{(n+2)(n+1)}{2},$$ we write

$$\dfrac{(n+2)(n+1)}{2}=28.$$

Multiplying both sides by 2, we obtain

$$(n+2)(n+1)=56.$$

Now we expand the product on the left:

$$n^2+3n+2=56.$$

Subtracting 56 from both sides gives

$$n^2+3n+2-56=0$$ $$\Rightarrow n^2+3n-54=0.$$

We look for factors of $$-54$$ whose sum is $$3.$$ The factor pair $$9$$ and $$-6$$ works, so

$$(n+9)(n-6)=0.$$

This yields two possible values $$n=-9$$ or $$n=6.$$ Since $$n$$ must be a non-negative integer, we take

$$n=6.$$

Now we need the sum of the coefficients of all terms in the expansion. A well-known fact is: “The sum of the coefficients in a multinomial expansion is obtained by substituting every variable by $$1.$$”

Here each term in $$\left(1-\dfrac{2}{x}+\dfrac{4}{y^2}\right)^6$$ contains the factors $$1,\;x^{-1},\;y^{-2}.$$ If we put $$x=1$$ and $$y=1,$$ every power of $$x$$ and $$y$$ becomes $$1,$$ leaving behind only the numerical coefficients. Thus

$$\text{Sum of coefficients}= \left(1-\dfrac{2}{1}+\dfrac{4}{1^2}\right)^6 = (1-2+4)^6 = (3)^6 = 729.$$

Hence, the correct answer is Option B.