To solve $^{33}C_2$, we use the formula for combinations:

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

Here, $n = 33$ and $r = 2$.

Substitute the values into the formula:

$^{33}C_2 = \frac{33!}{2!(33-2)!} = \frac{33 \cdot 32}{2 \cdot 1}$

Now, calculate step by step:

1. Multiply $33 \cdot 32$: $33 \cdot 32 = 1056$

2. Divide by $2$: $\frac{1056}{2} = 528$

Thus, $^{33}C_2 = 528$.

Final Answer: 528

Would you like further clarification or details on how this was calculated?

5 Related Questions:

1. How is the formula for combinations derived?
2. How do combinations differ from permutations?
3. What is the value of $^{10}C_3$?
4. Can you explain the difference between factorial and combination calculations?
5. What are some real-world applications of combinations?

Tip:

When calculating combinations with small $r$ values (like 2 or 3), it's often faster to expand the numerator directly without fully calculating the factorial!