document.write( "Question 1094921: You need to have a password with 4 letters followed by 4 even digits between 0 and 9, inclusive. If the characters and digits cannot be used more than once, how many choices do you have for your password? \n" ); document.write( "
Algebra.Com's Answer #709525 by math_helper(2461)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "There are 26 letters for the first letter choice.
\n" ); document.write( "25 for the 2nd
\n" ); document.write( "24 for the 3rd
\n" ); document.write( "23 for the 4th
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\n" ); document.write( "There are 5 choices for the first digit { 0, 2, 4, 6, or 8 }
\n" ); document.write( "4 for the 2nd
\n" ); document.write( "3 for the 3rd
\n" ); document.write( "2 for the 4th\r
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\n" ); document.write( "\n" ); document.write( "So there are 26*25*24*23*5*4*3*2 = 43056000 ways (=choices) to make a password this way.\r
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\n" ); document.write( "Alt solution: since order is important, we can use the permutation formula P(n,k)= n!/(n-k+1)!.
\n" ); document.write( "Viewing it this way, there are P(26,4) ways to pick (arrange) the letters and P(5,4) ways to pick (arrange) the digits: P(26,4)*P(5,4) = 358800*120 = 43056000.\r
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