To evaluate $^{12}C_2$, we use the combination formula:

$^{n}C_{r} = \frac{n!}{r!(n-r)!}$

Here, $n = 12$ and $r = 2$. Substituting these values into the formula:

$^{12}C_2 = \frac{12!}{2!(12-2)!} = \frac{12!}{2! \cdot 10!}$

Step 1: Simplify the factorials

Since $12! = 12 \cdot 11 \cdot 10!$, we can cancel the $10!$ in the numerator and denominator:

$^{12}C_2 = \frac{12 \cdot 11 \cdot 10!}{2! \cdot 10!}$

Cancel out $10!$:

$^{12}C_2 = \frac{12 \cdot 11}{2!}$

Step 2: Evaluate $2!$

We know $2! = 2$, so:

$^{12}C_2 = \frac{12 \cdot 11}{2}$

Step 3: Perform the calculation

$12 \cdot 11 = 132$ $\frac{132}{2} = 66$

Final Answer:

$^{12}C_2 = 66$

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Would you like a further explanation of combinations or another example?

Related Questions:

1. What is the formula for combinations and how is it derived?
2. How would you evaluate $^{10}C_3$?
3. How does a combination differ from a permutation?
4. What is the significance of combinations in probability and statistics?
5. How can we calculate combinations for larger numbers efficiently?

Tip:

In combinations, order does not matter, whereas in permutations, order matters. Always remember to distinguish between the two when solving problems!