document.write( "Question 891210: How many four-digit even numbers can be formed from the digits 0,2,3,5,6, and 9 if all for digits are different? The given answer is 108, but when i tried it out i got stuck because of the extra number and got 120 ..... \n" ); document.write( "
Algebra.Com's Answer #539767 by AnlytcPhil(1806)\"\" \"About 
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document.write( "There are two cases since 0 is an even digit and it can go last, but it\r\n" );
document.write( "can't go first.  So the number of choices for the first digit is different\r\n" );
document.write( "when we choose 0 for the last digit than if we choose 2 or 6 for the last\r\n" );
document.write( "digit.  So we must break this problem into two cases:\r\n" );
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document.write( "Case 1: 0 goes last\r\n" );
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document.write( "Choose the last (5th) digit as 0 in 1 way.\r\n" );
document.write( "Choose the 1st digit any of 5 ways. {2,3,5,6,9}\r\n" );
document.write( "Choose the 2nd digit any of the remaining 4 ways.\r\n" );
document.write( "Choose the 3rd digit any of the remaining 3 ways.\r\n" );
document.write( "Choose the 4th digit either of the remaining 2 ways. \r\n" );
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document.write( "That's 1󬊄󫢪 = 120 ways ending in 0\r\n" );
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document.write( "Case 2: 2 or 6 goes last:\r\n" );
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document.write( "Choose the last (5th) digit as in either of 2 ways.\r\n" );
document.write( "Choose the 1st digit any of 4 ways. (CANNOT choose 0)\r\n" );
document.write( "Choose the 2nd digit any of the remaining 4 ways. (CAN choose 0)\r\n" );
document.write( "Choose the 3rd digit any of the remaining 3 ways.\r\n" );
document.write( "Choose the 4th digit either of the remaining 2 ways.\r\n" );
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document.write( "That's 2󫶘󫢪 = 192 ways ending in 2 or 6\r\n" );
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document.write( "Total for the 2 cases = 120+192 = 312\r\n" );
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document.write( "Edwin
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