Key Concepts and Formulas

• Cartesian Product of Sets: For sets $A$ and $B$, the Cartesian product $A \times B$ is the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$. The cardinality of $A \times B$ is $|A \times B| = |A| \times |B|$.
• Combinations: The number of ways to choose $r$ objects from a set of $n$ distinct objects, without regard to order, is given by the binomial coefficient: $\binom{n}{r} = \frac{n!}{r!(n-r)!}$
• Combination Identity: $\binom{n}{r} = \binom{n}{n-r}$

Step-by-Step Solution

Step 1: Determine the Number of Elements in the Cartesian Product

We are given that $|A| = 5$ and $|B| = 2$. We need to find the number of elements in $A \times B$, which will be the universal set for our subsets. Using the formula for the size of a Cartesian product: $|A \times B| = |A| \times |B| = 5 \times 2 = 10$ Thus, $A \times B$ has 10 elements. Let $n = 10$.

Step 2: Identify the Required Subset Sizes

The problem states we need to find subsets of $A \times B$ with at least 3 and at most 6 elements. This means we are interested in subsets with sizes $r$ such that $3 \le r \le 6$. Therefore, we need to consider subsets with 3, 4, 5, and 6 elements.

Step 3: Calculate the Number of Subsets for Each Required Size

We will use the combination formula $\binom{n}{r}$ with $n=10$ for each value of $r$ identified in Step 2.

• Case 1: Subsets with 3 elements ($r=3$) The number of ways to choose 3 elements from 10 is: $\binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120$

• Case 2: Subsets with 4 elements ($r=4$) The number of ways to choose 4 elements from 10 is: $\binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 10 \times 3 \times 7 = 210$

• Case 3: Subsets with 5 elements ($r=5$) The number of ways to choose 5 elements from 10 is: $\binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 2 \times 9 \times 2 \times 7 = 252$

• Case 4: Subsets with 6 elements ($r=6$) The number of ways to choose 6 elements from 10 is: $\binom{10}{6} = \frac{10!}{6!(10-6)!} = \frac{10!}{6!4!}$ Using the identity $\binom{n}{r} = \binom{n}{n-r}$, we have $\binom{10}{6} = \binom{10}{10-6} = \binom{10}{4}$. Since we already calculated $\binom{10}{4}$ in Case 2: $\binom{10}{6} = 210$

Step 4: Sum the Number of Subsets

To find the total number of subsets satisfying the condition (at least 3 and at most 6 elements), we sum the results from each case: Total number of subsets = $\binom{10}{3} + \binom{10}{4} + \binom{10}{5} + \binom{10}{6} = 120 + 210 + 252 + 210 = 792$

Common Mistakes & Tips

• Carefully read the problem statement to correctly interpret the range for the number of elements in the subsets. "At least 3 and at most 6" means 3, 4, 5, and 6.
• Remember the combination formula and how to calculate factorials.
• Use the property $\binom{n}{r} = \binom{n}{n-r}$ to simplify calculations.

Summary

We found the number of elements in the Cartesian product $A \times B$ to be 10. We then calculated the number of subsets of $A \times B$ with 3, 4, 5, and 6 elements using combinations. Finally, we summed these values to find the total number of subsets satisfying the given condition, which is 792.

Final Answer

The final answer is \boxed{792}, which corresponds to option (D).