Let $c$ be the number of colors, which is 4.
Let $o$ be the number of options, which is 3.
Each car has a color and an option. The number of possible combinations of color and option is $c \times o = 4 \times 3 = 12$.
We want to find the largest number of cars we can guarantee to have the same color and option. This is a pigeonhole principle problem. The pigeonholes are the combinations of color and option, and the pigeons are the cars.
We have 12 pigeonholes (combinations of color and option) and 100000 cars (pigeons).
We can use the pigeonhole principle to find the minimum number of cars that must have the same color and option.
Let $k$ be the number of cars with the same color and option.
Then we have $\lceil \frac{100000}{12} \rceil = 8334$ cars with the same color and option.
To find the largest number of cars we can guarantee to have the same color and option, we divide the number of cars by the number of combinations of color and option and round up to the nearest integer.
$$ \left\lceil \frac{100000}{12} \right\rceil = 8334 $$
This means that if we have 100000 cars, we can guarantee that at least 8334 cars have the same color and option.
Therefore, the largest number of cars we can guarantee to have the same color and option is 8334.
The number of possible combinations of color and option is $4 \times 3 = 12$.
By the pigeonhole principle, if we have $n$ cars, the minimum number of cars with the same color and option is given by
$$ \left\lceil \frac{n}{12} \right\rceil $$
In our case, $n = 100000$, so the minimum number of cars with the same color and option is
$$ \left\lceil \frac{100000}{12} \right\rceil = 8334 $$
Therefore, we can guarantee to have at least 8334 cars with the same color and option.
Final Answer: The final answer is $\boxed{8334}$
