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 _________

We wish to count all three-digit even numbers that can be made with the six distinct digits $$\{0,1,3,4,6,7\}$$ when no digit is repeated.

A whole number is even exactly when its units (ones) digit is even. Inside our set the even digits are $$0,4,6.$$ Therefore the units place can be filled in three different ways, and we shall examine every one of those ways separately.

First let the units digit be $$0.$$

With $$0$$ fixed in the units place, the hundreds and tens places must be filled from the five remaining digits $$\{1,3,4,6,7\}.$$ For the hundreds place we may not use $$0$$ (because then the numeral would not be three-digit) and we may not repeat any digit. Hence there are

$$5$$ choices for the hundreds place.

After the hundreds position is chosen, one digit has been used up, so exactly $$4$$ digits are still free for the tens place.

Thus the number of numerals obtained in this sub-case is

$$1\;(\text{choice for units})\times5\times4=20.$$

Next let the units digit be $$4.$$

With $$4$$ already employed, the available digits for the hundreds place are $$\{0,1,3,6,7\}.$$ But $$0$$ cannot sit in the hundreds place, so we really have only $$4$$ admissible choices (namely $$1,3,6,7$$) for that position.

After deciding the hundreds digit, four digits are left, and all of them are now legal for the tens place (because the tens place may even be $$0$$). Therefore there are $$4$$ ways to choose the tens digit.

The count for this sub-case equals

$$1\;(\text{choice for units})\times4\times4=16.$$

Finally let the units digit be $$6.$$

The reasoning is identical to the previous case. The unused digits are $$\{0,1,3,4,7\},$$ of which $$0$$ cannot serve as the hundreds digit. Hence the hundreds place again admits $$4$$ choices, while the tens place can be filled in $$4$$ ways.

This yields

$$1\;(\text{choice for units})\times4\times4=16$$ different numerals.

We now add the results of the three mutually exclusive cases to obtain the total number of three-digit even numbers:

$$20+16+16=52.$$

So, the answer is $$52$$.