document.write( "Question 225343: The digits 2, 3, 4, 7 and 8 are each used once to form a five-digit number. What is the probability that the tens digit is odd and the number is divisible by 4? Express youranswer as a common fraction. \n" ); document.write( "
Algebra.Com's Answer #168171 by Edwin McCravy(20056)\"\" \"About 
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The digits 2, 3, 4, 7 and 8 are each used once to form a five-digit number. What is the probability that the tens digit is odd and the number is divisible by 4? Express your answer as a common fraction.
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document.write( "In order to have a success:\r\n" );
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document.write( "The tens digit must be a 3 or a 7 since it must be odd. \r\n" );
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document.write( "Any number divisible by 4 is such that the 2-digit number \r\n" );
document.write( "formed by the last two digits is divisible by 4.  The \r\n" );
document.write( "only 2-digit numbers beginning with 3 which are divisible\r\n" );
document.write( "by 4 are 32 and 36. The only 2-digit numbers beginning \r\n" );
document.write( "with 7 which are divisible by 4 are 72 and 76.  \r\n" );
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document.write( "Therefore,\r\n" );
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document.write( "there are only 4 types of numbers this can come out\r\n" );
document.write( "and be successful, they are:\r\n" );
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document.write( "1.    _ _ _ 3 2\r\n" );
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document.write( "or\r\n" );
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document.write( "2.    _ _ _ 3 6\r\n" );
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document.write( "or\r\n" );
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document.write( "3.    _ _ _ 7 2\r\n" );
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document.write( "or\r\n" );
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document.write( "4.    _ _ _ 7 6\r\n" );
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document.write( "For any one of these four we can choose the first digit\r\n" );
document.write( "any of the remaining 3 ways, the 2nd digit any of the\r\n" );
document.write( "then remaining 2 ways, and there will be only one remaining\r\n" );
document.write( "choice for the 3rd digit.  That's 3! for each of the 4 types,\r\n" );
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document.write( "So the numerator is 4*3!\r\n" );
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document.write( "The denominator is the number of ways of choosing ANY 5-digit\r\n" );
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document.write( "We may choose the 1st digit any of 5 ways, the 2nd digit any\r\n" );
document.write( "of of the 4 remaining digits, the 3rd digit any of the 3 \r\n" );
document.write( "remaining digits, the 4th digit any of the 2 remaining digits\r\n" );
document.write( "and the 5th is the 1 remaining digit.\r\n" );
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document.write( "So the denominator is 5*4*3*2*1 or 5!\r\n" );
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document.write( "Answer = \r\n" );
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document.write( "Edwin
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