Animated Solution for Mathematics - Permutations and Combinations: If the number of five digit numbers with distinct digits and 2 at the 10th place is $336k$, then $k$ is equal to:

Imagine you are standing before a grand, five-digit vault. To open it, you must select a unique combination of digits, but there are strict mathematical rules that govern this process. This is the essence of combinatorics, the study of counting possibilities.

Visualize our five-digit number as five empty slots: $[Ten-Thousands][Thousands][Hundreds][Tens][Units]$. We have ten digits available: $\{0,1,2,3,4,5,6,7,8,9\}$. The problem imposes two constraints: all digits must be distinct, and the digit $2$ is permanently locked into the tens place.

In combinatorics, always start with the most restrictive condition. Here, that is the tens place. Since the digit $2$ is fixed, we have only $1$ choice for this slot.

Now, look at the ten-thousands place. This is a classic trap! We cannot place a $0$ here, or our 5-digit number collapses into a 4-digit number. Furthermore, we cannot use the digit $2$ because it is already taken. Out of our ten digits, we exclude $0$ and $2$, leaving us with $8$ choices for the first slot.

Next, consider the thousands place. The restriction on $0$ is lifted, but we have used two digits already: the $2$ in the tens place and one digit in the ten-thousands place. Thus, we have $10-2=8$ choices remaining.

For the hundreds place, we have used three digits, leaving $10-3=7$ choices. Finally, for the units place, we have used four digits, leaving $10-4=6$ choices.

By the Fundamental Principle of Counting, we multiply these choices:

Let us calculate the product:

We have found that there are $2688$ such numbers. The problem states that this total is equal to $336k$. So, we set up the equation:

Solving for $k$, we find:

The vault opens. The logic holds. You have mastered the constraints, and the final value is $k=8$.