Animated Solution for Mathematics - Permutations and Combinations: The number of three-digit even numbers, formed by the digits 0, 1, 3, 4, 6, 7 if the repetition of digits is not allowed, is .

The objective is to form three-digit even numbers using the set of digits $S=\{0,1,3,4,6,7\}$ without repetition. A three-digit number cannot have $0$ in the hundreds place.

Because $0$ is an even digit, it influences the units place differently than the other even digits ($4$ and $6$). We must resolve this dependency by splitting the problem into two mutually exclusive cases.

The Zero at the End

In this scenario, we fix the units place as $0$. There is exactly $1$ way to fill this position.

Since $0$ is already occupied, the hundreds place can be filled by any of the remaining $5$ digits $\{1,3,4,6,7\}$. There are $5$ ways to fill the hundreds place.

For the tens place, we have used two digits (one at the units, one at the hundreds). Out of the original $6$ digits, $4$ remain available.

By the Fundamental Principle of Counting, the total for this case is:

The Non-Zero Even Digits

In this scenario, the units place must be either $4$ or $6$. There are $2$ ways to fill the units place.

When we move to the hundreds place, we must exclude both the digit used in the units place and the digit $0$. Out of the $6$ original digits, $2$ are now unavailable for the hundreds place.

This leaves $4$ valid choices for the hundreds place.

For the tens place, we have used two digits (one at the units, one at the hundreds). Out of the original $6$ digits, $4$ remain available. Note that $0$ is now a valid choice for the tens place.

The total for this case is:

We have accounted for the special behavior of $0$ by splitting our logic into two distinct, mutually exclusive scenarios. To find the final answer, we sum the results of both cases:

The total number of three-digit even numbers that can be formed is 52.