document.write( "Question 35052: Solve using whatever counting methods are necessary. Please show work.\r
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document.write( "How many 3 digit numbers are possible using 2, 3, 5, or 7 if you can repeat digits?\r
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document.write( "Any help would be greatly appreciated! Thank you!
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document.write( "Smiles! \n" );
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Algebra.Com's Answer #21252 by longjonsilver(2297)![]() ![]() You can put this solution on YOUR website! i answered this yesterday. Have a look at my recent answers.\r \n" ); document.write( "\n" ); document.write( "Answer is 24: 4*3*2 \n" ); document.write( "================================================================ \n" ); document.write( "Apologies, i did not read the question properly and assumed it was the same as before.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "This is actually easier than the other example though.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "With repetition we can have numbers like 222 or 353 etc.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "First position: How many numbers are allowed there? Answer is 4 \n" ); document.write( "Second position: How many numbers are allowed there? Answer is again 4 \n" ); document.write( "Third position: How many numbers are allowed there? Answer is again 4\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "So in total we have 4*4*4 permutations --> 64.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Jon \n" ); document.write( " \n" ); document.write( " |