document.write( "Question 900518: there are 4 dimes, 3 nickels, and 2 quarters. \r
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Algebra.Com's Answer #546472 by Edwin McCravy(20055) $\"\"$ $\"About$

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There are 4 dimes, 3 nickels, and 2 quarters. \r
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\n" ); document.write( "made so that the value of the coins is at least 25 cents.
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document.write( "We will get the number of possible selections, and then subtract \r\n" );
document.write( "the number less than 25 cents.\r\n" );
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document.write( "We can choose the number of dimes 5 ways 0,1,2,3, or 4.\r\n" );
document.write( "We can choose the number of nickels 4 ways 0,1,2, or 3.\r\n" );
document.write( "We can choose the number of quarters 3 ways 0,1, or 2.\r\n" );
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document.write( "That's 5*4*3 = 60 selections.\r\n" );
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document.write( "Now we must subtract from the 60 the number of selections \r\n" );
document.write( "of coins that are less than 25 cents.  These will involve only\r\n" );
document.write( "dimes and nickels.\r\n" );
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document.write( "To get a selection of coins worth less than 25 cents:\r\n" );
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document.write( "If we use no dimes, we can use 0,1,2, or all 3 nickels.\r\n" );
document.write( "That's 4 selections less than 25 cents.  (That includes\r\n" );
document.write( "the choice of NO coins at all, which is counted as a choice \r\n" );
document.write( "contained in the 60, which we must subtract!)\r\n" );
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document.write( "If we use exactly 1 dime, we can use 0,1, or 2 nickels.\r\n" );
document.write( "That's 3 combinations less than 25 cents.\r\n" );
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document.write( "And there is 1 other selection less than 25 cents, 2 dimes \r\n" );
document.write( "and no nickels.\r\n" );
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document.write( "So that's 4+3+1 = 8 selections which we must subtract from the 60.\r\n" );
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document.write( "Answer: 60-8 = 52 selections of coins worth 25 cents or more.  \r\n" );
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document.write( "Edwin

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