1. Factorise the First Base Expression
We begin by simplifying the first expression inside the exponent using grouping:
$$1 + x + y + \frac{x}{y} = (1 + x) + y\left(1 + \frac{1}{y}\right)$$
This layout is easier to handle if rearranged differently to isolate common binomials:
$$1 + \frac{x}{y} + x + y = \frac{y + x}{y} + (x + y)$$
Factoring out the shared numerator $$(x + y)$$:
$$(x + y)\left(\frac{1}{y} + 1\right) = (x + y)\left(1 + \frac{1}{y}\right)$$
Substituting this product back into the power of $$10$$ gives the first simplified component:
$$f_1(x, y) = (x + y)^{10} \left(1 + \frac{1}{y}\right)^{10}$$
2. Factorise the Second Base Expression
Now we apply a similar grouping approach to the second expression:
$$1 + y + \frac{1}{x} + \frac{y}{x} = (1 + y) + \frac{1}{x}(1 + y)$$
Factoring out the shared expression $$(1 + y)$$:
$$(1 + y)\left(1 + \frac{1}{x}\right)$$
Substituting this back into the power of $$10$$ gives the second simplified component:
$$f_2(x, y) = (1 + y)^{10} \left(1 + \frac{1}{x}\right)^{10}$$
3. Determine the Unique Exponent Pairs for Each Component
We analyze the exponent footprints $$(a, b)$$ of the terms generated by each expanded part independently.
For the first component:
$$f_1(x, y) = \left( \sum_{i=0}^{10} \binom{10}{i} x^{10-i} y^i \right) \left( \sum_{j=0}^{10} \binom{10}{j} y^{-j} \right) = \sum_{i=0}^{10} \sum_{j=0}^{10} \binom{10}{i} \binom{10}{j} x^{10-i} y^{i-j}$$
The exponent pairs generated are:
$$a = 10 - i \quad \text{and} \quad b = i - j$$
Since $$0 \leq i \leq 10$$ and $$0 \leq j \leq 10$$, each choice of the pair $$(i, j)$$ yields a unique set of exponents. The total number of distinct terms from the first component is:
$$11 \times 11 = 121 \text{ terms}$$
For the second component:
$$f_2(x, y) = \left( \sum_{k=0}^{10} \binom{10}{k} y^k \right) \left( \sum_{m=0}^{10} \binom{10}{m} x^{-m} \right) = \sum_{k=0}^{10} \sum_{m=0}^{10} \binom{10}{k} \binom{10}{m} x^{-m} y^k$$
The exponent pairs generated are:
$$a = -m \quad \text{and} \quad b = k$$
Since $$0 \leq k \leq 10$$ and $$0 \leq m \leq 10$$, every pair $$(k, m)$$ produces a unique set of exponents. The total number of distinct terms from the second component is:
$$11 \times 11 = 121 \text{ terms}$$
4. Find and Subtract the Overlapping Terms
When combining the two expansions, terms with identical exponent pairs $$(a, b)$$ will merge together. To find the true number of unique terms, we use the Principle of Inclusion-Exclusion:
$$\text{Total Terms } T = \text{Terms}(f_1) + \text{Terms}(f_2) - \text{Terms}(f_1 \cap f_2)$$
Let's find when an exponent pair $$(a, b)$$ belongs to both expansions simultaneously:
From the first expansion: $$a = 10 - i \geq 0$$ (since $$i \leq 10$$).
From the second expansion: $$a = -m \leq 0$$ (since $$m \geq 0$$).
For a term to exist in both, its $$x$$-exponent $$a$$ must satisfy both conditions:
$$a \geq 0 \quad \text{and} \quad a \leq 0 \implies a = 0$$
Setting $$a = 0$$ determines the exact matching indices for both expansions:
- In the first expansion: $$10 - i = 0 \implies i = 10$$.
- In the second expansion: $$-m = 0 \implies m = 0$$.
Now we analyze the corresponding $$y$$-exponent $$b$$ under these index constraints:
- From the first expansion with $$i = 10$$: $$b = 10 - j$$. Since $$0 \leq j \leq 10$$, this produces $$b \in \{0, 1, 2, \dots, 10\}$$.
- From the second expansion with $$m = 0$$: $$b = k$$. Since $$0 \leq k \leq 10$$, this also produces $$b \in \{0, 1, 2, \dots, 10\}$$.
The overlapping pairs are precisely the $$11$$ distinct terms where $$a = 0$$ and $$0 \leq b \leq 10$$:
$$\text{Overlapping Terms} = 11$$
5. Calculate the Final Value
Applying these values to our inclusion-exclusion formula:
$$T = 121 + 121 - 11$$
$$T = 242 - 11 = 231$$
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