document.write( "Question 1088889: Find the probability that the combination of 4 letters will be in the order EAFC. The same letter can be used more than once.
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\n" ); document.write( "b. 1/ 358,800
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\n" ); document.write( "c. 1/ 2,500
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\n" ); document.write( "d. 1/ 104 \n" ); document.write( "
Algebra.Com's Answer #703260 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "There are 26 ways the first letter can be.\r\n" );
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document.write( "For each of those 26 ways, there are 26 ways the second letter can be.\r\n" );
document.write( "That's 26×26 ways the first and second letters can be.\r\n" );
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document.write( "For each of those 26×26 ways, there are 26 ways the third letter can be.\r\n" );
document.write( "That's 26×26×26 ways the first, second and third letters can be.\r\n" );
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document.write( "For each of those 26×26×26 ways, there are 26 ways the last letter can be.\r\n" );
document.write( "That's 26×26×26×26 ways the first, second, third and last letters can be.\r\n" );
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document.write( "So we multiply 26×26×26×26 and get 456,976.\r\n" );
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document.write( "So there's just 1 way to get EAFC out of all those 456,976.\r\n" );
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document.write( "So the probability is 1 out of 456,876 or 1/456,976\r\n" );
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document.write( "Edwin
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