The expression \(^nC_2\) represents the number of ways to choose 2 items from a set of \(n\) distinct items. The formula for combinations is:

\[
^nC_r = \frac{n!}{r!(n-r)!}
\]

For \(^nC_2\):

\[
^nC_2 = \frac{n!}{2!(n-2)!}
\]

Simplifying this, we get:

\[
^nC_2 = \frac{n \times (n-1) \times (n-2)!}{2 \times 1 \times (n-2)!}
\]

The \((n-2)!\) terms cancel out, leaving:

\[
^nC_2 = \frac{n \times (n-1)}{2}
\]

So, \(^nC_2\) is calculated by taking the product of \(n\) and \(n-1\) and then dividing by 2.

For example:
- If \(n = 5\), then \(^5C_2 = \frac{5 \times 4}{2} = 10\).
- If \(n = 6\), then \(^6C_2 = \frac{6 \times 5}{2} = 15\).

Would you like further clarification on how to calculate or apply combinations?

Here are 5 related questions:
1. How do you derive the formula for \(^nC_r\)?
2. What is the significance of \(^nC_2\) in combinatorial problems?
3. How does \(^nC_2\) relate to the sum of the first \(n-1\) integers?
4. Can you explain the difference between \(^nC_2\) and \(^nP_2\)?
5. How would you use \(^nC_2\) in a probability scenario involving pairs?

**Tip:** To quickly calculate combinations like \(^nC_2\), remember that you're essentially finding the number of unique pairs that can be formed from \(n\) items.

Explore how to calculate n choose 2 (nC2) combinations using the factorial formula. Learn the significance of nC2 in combinatorial problems and its applications in probability scenarios involving pairs. Examples provided for better understanding.

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