<s>[INST] How many even numbers of $3$ digits, between $400$ and $700$ can be formed if repeating digits are not allowed? I have this statement:

How many even numbers of $3$ digits, between $400$ and $700$ can be formed if repeating digits are not allowed?

My attempt was:
In the case that the number start in $4$, so i have $8 * 4 = 32$ ways
In the case that the number start in $5$, so i have $8 * 5 = 40$ ways
In the case that the number start in $6$, so i have $8 * 4 = 32$ ways
In the case that the numbers start in $7$ only can be $700$, so is $1$ way.
Therefore, the total ways are $105$
But the correct answer is $104$ according to the guide.
What is wrong with my development?
 [/INST] The digits in $700$ are not distinct, so you should not have counted it.
Since the digits of $700$ are not distinct, the leading digit of a number with distinct digits must be 4, 5, or 6.
If the units digit is 4 or 6, there are two choices for the hundreds digit since it must differ from the units digit and eight choices for the tens digit since it must differ from both the units digits and the hundreds digit.  Hence, there are $2 \cdot 2 \cdot 8$ such numbers.
If the units digit is 0, 2, or 8, there are three choices for the hundreds digit since 4, 5, 6 are all available and eight choices for the the tens digit since it must differ from both the units digit and the hundreds digit.  Thus, there are $3 \cdot 3 \cdot 8$ such numbers.
In total, there are $2 \cdot 2 \cdot 8 + 3 \cdot 3 \cdot 8 = 32 + 72 = 104$ even numbers between $400$ and $700$ with distinct digits.
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