We will evaluate the given expression step by step. The expression is:

$\binom{12}{3} \cdot \binom{20}{2} + \binom{12}{4} \cdot \binom{20}{2} + \binom{12}{5} \cdot \binom{20}{0}$

Step 1: Recall the formula for combinations

The formula for combinations is: $\binom{n}{r} = \frac{n!}{r!(n-r)!}$

Step 2: Calculate each term

1. First term: $\binom{12}{3} \cdot \binom{20}{2}$ $\binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12 \cdot 11 \cdot 10}{3 \cdot 2 \cdot 1} = 220$ $\binom{20}{2} = \frac{20!}{2!(20-2)!} = \frac{20 \cdot 19}{2 \cdot 1} = 190$ $\binom{12}{3} \cdot \binom{20}{2} = 220 \cdot 190 = 41800$

2. Second term: $\binom{12}{4} \cdot \binom{20}{2}$ $\binom{12}{4} = \frac{12!}{4!(12-4)!} = \frac{12 \cdot 11 \cdot 10 \cdot 9}{4 \cdot 3 \cdot 2 \cdot 1} = 495$ $\binom{20}{2} = 190 \quad \text{(calculated earlier)}$ $\binom{12}{4} \cdot \binom{20}{2} = 495 \cdot 190 = 94050$

3. Third term: $\binom{12}{5} \cdot \binom{20}{0}$ $\binom{12}{5} = \frac{12!}{5!(12-5)!} = \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 792$ \binom{20}{0} = 1 \quad \text{(since \binom{n}{0} = 1$for any$n)} $\binom{12}{5} \cdot \binom{20}{0} = 792 \cdot 1 = 792$

Step 3: Add the results

Now, sum up the three terms: $41800 + 94050 + 792 = 136642$

Final Answer:

$\boxed{136642}$

Would you like a deeper explanation of any step, or additional examples? Here are 5 related questions to expand on this:

1. How do combinations differ from permutations?
2. What is the general formula for calculating $\binom{n}{r}$?
3. How would the result change if the numbers in the problem were larger (e.g., $\binom{15}{5}$)?
4. Can you explain the connection between binomial coefficients and Pascal's Triangle?
5. How are combinations used in probability problems?

Tip: For faster calculations of combinations, try using factorial simplifications and cancel terms early.