Question 1190087
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A 4-digit number is selected from the numbers {1,2,3,4,5,6} where the digits are selected without replacement.
How many 4-digit numbers can be chosen that are even and greater than 4000?
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            The traditional and the standard formulation and the meaning of the problem is that the digits are used without repetition . . . 



In this problem,  there are two restrictions:


            (a)    the first (mostleft, thousands) digit must be 4, or 5, or 6;


            (b)    the last digit  (ones digit)  must be  2,  or  4,  or  6  in order for the number be even.


These restrictions are not independent; therefore, an accurate analysis is required.



<pre>
Case 1.  Let the last digit be 2.


         Then the leading digit can be 4 or 5 or 6, giving 3 options;

         the second digit can be any of remaining 6-2 = 4 digits, giving 4 options;

         the third digit can be any of remaining 6-3 = 3 digits, giving 3 options.


         Thus in this case we have 3 * 4 * 3 = 36 possible 4-digit numbers.



Case 2.  Let the last digit be 4.

         Then the leading digit can be 5 or 6, giving 2 options;

         the second digit can be any of remaining 6-2 = 4 digits, giving 4 options;

         the third digit can be any of remaining 6-3 = 3 digits, giving 3 options.


         Thus in this case we have 2 * 4 * 3 = 24 possible 4-digit numbers.



Case 3.  Let the last digit be 6.     (In this case the analysis is similar to case 2).

         Then the leading digit can be 4 or 5, giving 2 options;

         the second digit can be any of remaining 6-2 = 4 digits, giving 4 options;

         the third digit can be any of remaining 6-3 = 3 digits, giving 3 options.


         Thus in this case we have 2 * 4 * 3 = 24 possible 4-digit numbers.



Thus the total of cases 1, 2 and 3  is  24 + 36 + 24 = 84 possible 4-digit numbers.


<U>ANSWER</U>.  There are 84 4-digit numbers under imposed conditions.
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