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By our fundamental counting principle, we can make 2 * 3 * 4 different combinations of dimes, nickles, and pennies (you may think 1*2*3, but remember there's actually one extra way for each coin and that is to have 0 coins).\r
\n" ); document.write( "\n" ); document.write( "So there are 24 different combinations. Now do any of these coins yield the same amount? Notice that 2 nickels = dime. Notice that there are eight cases where we have either a certain number of pennies (0-3) + dime or a certain number of pennies (0-3) + 2 nickels. So, we need to eliminate either the dime case or the 2 nickels case. Either way gives us 4 less ways. And so we can have 20 distinct amounts for our coins. If your teacher does not consider the all 0 case to be an amount, then there are really only 19.\r
\n" ); document.write( "\n" ); document.write( "These amounts being\r
\n" ); document.write( "\n" ); document.write( "(0)
\n" ); document.write( "1
\n" ); document.write( "2
\n" ); document.write( "3
\n" ); document.write( "5
\n" ); document.write( "6
\n" ); document.write( "7
\n" ); document.write( "8
\n" ); document.write( "10
\n" ); document.write( "11
\n" ); document.write( "12
\n" ); document.write( "13
\n" ); document.write( "15
\n" ); document.write( "16
\n" ); document.write( "17
\n" ); document.write( "18
\n" ); document.write( "20
\n" ); document.write( "21
\n" ); document.write( "22
\n" ); document.write( "23
\n" ); document.write( "-----
\n" ); document.write( "19 or 20 different amounts depending on 0 case \n" ); document.write( "