document.write( "Question 35052: Solve using whatever counting methods are necessary. Please show work.\r
\n" ); document.write( "\n" ); 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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Algebra.Com's Answer #21252 by longjonsilver(2297)\"\" \"About 
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
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\n" ); document.write( "Apologies, i did not read the question properly and assumed it was the same as before.\r
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\n" ); document.write( "\n" ); document.write( "This is actually easier than the other example though.\r
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\n" ); document.write( "\n" ); document.write( "With repetition we can have numbers like 222 or 353 etc.\r
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\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
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\n" ); document.write( "\n" ); document.write( "So in total we have 4*4*4 permutations --> 64.\r
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\n" ); document.write( "\n" ); document.write( "Jon
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