Key Concepts and Formulas

• Fundamental Principle of Counting (Multiplication Principle): If there are $m$ ways to do one thing and $n$ ways to do another, then there are $m \times n$ ways to do both.
• Permutations: Arrangements of objects in a specific order. The number of permutations of $n$ distinct objects taken $r$ at a time is denoted by $P(n, r)$ or $_nP_r$ and is given by $\frac{n!}{(n-r)!}$.
• Digits: The digits are $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.

Step-by-Step Solution

Step 1: Representing the Five-Digit Number

We represent the five-digit number as five blanks: $\underline{\hspace{0.8cm}} \quad \underline{\hspace{0.8cm}} \quad \underline{\hspace{0.8cm}} \quad \underline{\hspace{0.8cm}} \quad \underline{\hspace{0.8cm}}$

Step 2: Fixing the Tens Place

The digit in the tens place is fixed as 2. This leaves us with one choice for this position. $\underline{\hspace{0.8cm}} \quad \underline{\hspace{0.8cm}} \quad \underline{\hspace{0.8cm}} \quad \mathbf{2} \quad \underline{\hspace{0.8cm}}$ Number of choices for the tens place = 1. Digits used so far: {2}

Step 3: Filling the Ten-Thousands Place

The ten-thousands place cannot be 0 and cannot be 2 (since all digits must be distinct). Therefore, we have 8 choices (1, 3, 4, 5, 6, 7, 8, 9). $\mathbf{8} \quad \underline{\hspace{0.8cm}} \quad \underline{\hspace{0.8cm}} \quad \mathbf{2} \quad \underline{\hspace{0.8cm}}$ Number of choices for the ten-thousands place = 8. Digits used so far: {2, one other digit}

Step 4: Filling the Thousands Place

The thousands place can be 0 but cannot be 2 and cannot be the digit chosen for the ten-thousands place. So we have $10 - 2 = 8$ choices. $\mathbf{8} \quad \mathbf{8} \quad \underline{\hspace{0.8cm}} \quad \mathbf{2} \quad \underline{\hspace{0.8cm}}$ Number of choices for the thousands place = 8. Digits used so far: {2, two other distinct digits}

Step 5: Filling the Hundreds Place

The hundreds place cannot be 2 and cannot be any of the two digits already used in the ten-thousands and thousands places. So we have $10 - 3 = 7$ choices. $\mathbf{8} \quad \mathbf{8} \quad \mathbf{7} \quad \mathbf{2} \quad \underline{\hspace{0.8cm}}$ Number of choices for the hundreds place = 7. Digits used so far: {2, three other distinct digits}

Step 6: Filling the Units Place

The units place cannot be 2 and cannot be any of the three digits already used in the ten-thousands, thousands and hundreds places. So we have $10 - 4 = 6$ choices. $\mathbf{8} \quad \mathbf{8} \quad \mathbf{7} \quad \mathbf{2} \quad \mathbf{6}$ Number of choices for the units place = 6. Digits used so far: {2, four other distinct digits}

Step 7: Calculating the Total Number of Five-Digit Numbers

Using the Fundamental Principle of Counting, the total number of such five-digit numbers is: $8 \times 8 \times 7 \times 1 \times 6 = 2688$

Step 8: Solving for k

We are given that the number of such five-digit numbers is $336k$. Therefore: $336k = 2688$ $k = \frac{2688}{336} = 8$

Common Mistakes & Tips

• Forgetting the zero restriction: Be mindful that the first digit of a multi-digit number cannot be zero.
• Not accounting for distinct digits: Ensure that you are removing already used digits when calculating the number of options for each position.
• Misapplying the counting principle: The fundamental principle of counting is only applicable when events are independent. Verify the independence of each step.

Summary

We found the total number of five-digit numbers with distinct digits, with 2 in the tens place, by considering the restrictions on each digit and applying the fundamental principle of counting. We determined that the total number of such numbers is 2688, and by equating this to 336k, we found that k = 8.

Final Answer

The final answer is \boxed{8}, which corresponds to option (B).