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How many 4-digit even numbers can be formed from the digits 0 to 9
if each digit is to be used only once in each number ?
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The fact that the number is an EVEN number means that the last digit is one of 5 even digits 0, 2, 4, 6, or 8.
In my solution, I will consider two cases separately:
case (a): the last digit is 0 (zero),
and
case (b): the last digit is any of the remaining 4 even digits 2, 4, 6 or 8.
Case (a): the last digit is 0 (zero)
Then the first (most-left) digit is any of 9 remaining digits;
the second digit is any of remaining 8 digits;
the third digit is any of remaining 7 digits.
So, the total number of possible options is 9*8*7 = 504 in this case.
Case (b): the last digit is any of remaining 4 digits 2, 4, 6 or 8.
Then the first (most-left) digit is any of 8 remaining digits (keep in mind that the leading digit CAN NOT be 0 (!));
the second digit is any of 8 remaining digits (zero is ALLOWED in this position);
the third digit is any of 7 remaining digits (zero is ALLOWED in this position).
So, the total number of possible options is 8*8*7 = 448 in this case.
Thus the total number of possibilities is 504 + 448 = 952.
ANSWER. 952 four-digit even numbers can be formed from the digits 0 to 9 if each digit is to be used only once in each number.
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Solved.
The major lesson to learn from my solution is splitting the analysis in two cases.