document.write( "Question 1177880: How many 4 digit numbers can be formed from the digits 0-9 \n" ); document.write( "

Algebra.Com's Answer #807010 by MathLover1(20850) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( " if repetitions are allowed,\r
\n" ); document.write( "\n" ); document.write( "Two ways to get the answer:\r
\n" ); document.write( "\n" ); document.write( "1. There are $\"+9999\"$ integers starting with $\"1+\"$ and ending with $\"9999\"$ .
\n" ); document.write( "But the $\"999\"$ integers starting with $\"1+\"$ and ending with $\"999\"$ have less than $\"4+\"$ digits,\r
\n" ); document.write( "\n" ); document.write( "so the desired number is $\"9999-999\"$ or $\"9000\"$ ways.\r
\n" ); document.write( "\n" ); document.write( "2.
\n" ); document.write( "there is total of $\"10\"$ digits from $\"0\"$ to $\"9\"$
\n" ); document.write( "so, there is $\"10\"$ ways to choose each ones, tens, hundredths, and thousands'
\n" ); document.write( "that is $\"10%5E4\"$
\n" ); document.write( "since $\"0+\"$ cannot be in place of thousands', only ones, tens, hundredths which is $\"10%5E3\"$ three digit numbers , deduct it\r
\n" ); document.write( "\n" ); document.write( " $\"10%5E4-10%5E3=10000-1000=9000\"$ \r
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