Question 1170640
.
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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<pre>
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.
</pre>


<U>Case (a)</U>: &nbsp;&nbsp;the last digit is &nbsp;0 &nbsp;(zero)


<pre>
    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.
</pre>


<U>Case (b)</U>: &nbsp;&nbsp;the last digit is any of remaining &nbsp;4 &nbsp;digits &nbsp;2, 4, 6 or 8.


<pre>
    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.
</pre>

Thus the total number of possibilities is 504 + 448 = 952.



<U>ANSWER</U>.  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.