The image contains two combination problems, and the combinations are being calculated using both the combination formula and Pascal's triangle.

The formula for calculating combinations is:

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

where $n$ is the total number of items, and $r$ is the number of items to choose.

You are also provided with a Pascal's triangle, which helps simplify these calculations. For the combination values, you can find them in Pascal’s triangle by looking at row $n$ and selecting the $r$-th element.

Problems:

1. $^3C_2$
2. $^3C_3$

Solving using Pascal's Triangle:

• The third row of Pascal’s triangle (starting from row 0) is: $1, 3, 3, 1$
• $^3C_2$ is the third element in this row, which is 3.
• $^3C_3$ is the fourth element in this row, which is 1.

Conclusion:

1. $^3C_2 = 3$
2. $^3C_3 = 1$

Would you like further details or explanations on this? Here are five related questions to expand this:

1. What is the general use of Pascal’s Triangle in combinatorics?
2. How can you derive the combination formula from Pascal’s Triangle?
3. What is the significance of symmetry in Pascal’s Triangle?
4. How would you calculate $^4C_2$ using the combination formula?
5. Can you use Pascal's Triangle to calculate powers of binomials?

Tip: Pascal's Triangle is an efficient tool for quickly calculating combinations and also for expanding binomials in algebra.