1. The problem is to evaluate the expression involving combinations: $$8C_0 \cdot 12C_4 + 8C_1 \cdot 12C_3 + 8C_2 \cdot 12C_2$$ which represents selecting groups from two sets. 2. Recall the combination formula: $$nC_r = \frac{n!}{r!(n-r)!}$$ where $n!$ is the factorial of $n$. 3. Calculate each term: - $8C_0 = 1$ (choosing 0 from 8) - $12C_4 = \frac{12!}{4!8!} = 495$ - $8C_1 = 8$ - $12C_3 = \frac{12!}{3!9!} = 220$ - $8C_2 = \frac{8!}{2!6!} = 28$ - $12C_2 = \frac{12!}{2!10!} = 66$ 4. Multiply and sum the terms: $$1 \times 495 + 8 \times 220 + 28 \times 66 = 495 + 1760 + 1848 = 4103$$ 5. This sum, 4103, represents the total number of ways to select the groups as described by the problem. This step-by-step approach shows how to use combinations and factorials to solve the problem clearly and accurately.