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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.

How many subsets of size 2 does the set S = {1, 2, 3, 4, 5, 6} have?

Learn how to calculate how many subsets of size 2 can be formed from the set S = {1, 2, 3, 4, 5, 6}. This problem covers key combinatorics concepts, including the combination formula. Suitable for students in Grades 9-11.

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