The question asks how many subsets of size 2 can be formed from the set $S = \{1, 2, 3, 4, 5, 6\}$.

To solve this, we need to use the concept of combinations, since the order of the elements in the subset doesn't matter.

The formula for combinations is:

$\binom{n}{k} = \frac{n!}{k!(n-k)!}$

Where:

• $n$ is the total number of elements in the set $S$,
• $k$ is the size of the subset we want to form.

Here, $n = 6$ and $k = 2$. Applying the formula:

$\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15$

So, there are 15 subsets of size 2 that can be formed from the set $S = \{1, 2, 3, 4, 5, 6\}$.

Would you like more details on this process or have any questions?

Here are 5 related questions:

1. How many subsets of size 3 can be formed from the set $S = \{1, 2, 3, 4, 5, 6\}$?
2. What is the total number of subsets that can be formed from a set with 6 elements?
3. How do combinations differ from permutations in set theory?
4. What is the formula for the number of subsets of any size from a set of size $n$?
5. Can the concept of combinations be applied to other areas outside of set theory?

Tip: Combinations are used when the order of selection does not matter, while permutations are used when the order matters.