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.