document.write( "Question 926887: How many 3-number combinations might a student have to try in order to open a lock if the numbers from 0 to 39 appear on the dial? Assume that the second number may be the same as the first (since the dial will be rotated in a different direction), but the third number must be at least 5 away from the second. \n" ); document.write( "
Algebra.Com's Answer #563195 by Edwin McCravy(20060)\"\" \"About 
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document.write( "There are 40 ways to choose the first number\r\n" );
document.write( "and there are 40 ways to choose the second number.\r\n" );
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document.write( "Now we take some examples to determine how many choices there are for \r\n" );
document.write( "the third number \r\n" );
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document.write( "If the second number is 23, then the 3rd number cannot be any of these\r\n" );
document.write( "9 numbers 19,20,21,22,23,24,25,26,or 27, since they are not at least 5 \r\n" );
document.write( "numbers away from 23.  Same for all other 40 second numbers. \r\n" );
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document.write( "Let's look at a second example which includes some on both sides of 0.\r\n" );
document.write( "If second number is 2, we couldn't choose any of the 9 numbers 38,39,0,1,2,3,4,5,6.\r\n" );
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document.write( "So thats 40-9 or 31 choices for the 3rd number.\r\n" );
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document.write( "Answer 40*40*31 = 49600 ways.\r\n" );
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document.write( "Edwin

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