document.write( "Question 1139410: From the letters of the word FOREVER, 4-letter codes were formed. How many codes:
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\n" ); document.write( "(d) did not have any repeated letters?
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Algebra.Com's Answer #760082 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
From the letters of the word FOREVER, 4-letter codes were formed. How many codes:
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Case 1: No repeated letters. These are 5 letters (FOREV) PERMUTE 4.\r\n" );
document.write( "That's 5P4 = 5∙4∙3∙2 = 120\r\n" );
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document.write( "Case 2: With R repeated but not E: \r\n" );
document.write( "Choose the 2 positions for the R's to go in 4C2 = 6 ways.\r\n" );
document.write( "That leaves 2 unfilled positions.\r\n" );
document.write( "Choose the letter to go in the leftmost unfilled position from FOEV\r\n" );
document.write( "4 ways.\r\n" );
document.write( "That leaves only 1 unfilled position.\r\n" );
document.write( "Choose the letter to go in it in 3 ways.\r\n" );
document.write( "That's 6∙4∙3 = 72\r\n" );
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document.write( "Case 3: With E repeated but not R: \r\n" );
document.write( "Same as Case 2.  72 ways \r\n" );
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document.write( "Case 4: with 2 R's and 2 E's.\r\n" );
document.write( "Choose the 2 positions for the R's in 4C2 = 6 ways.\r\n" );
document.write( "Fill the remaining 2 positions with E's in 1 way.\r\n" );
document.write( "That's 6∙1 = 6 ways for case 3.\r\n" );
document.write( "Total from all three cases: 120+72+72+6 = 270 ways. \r\n" );
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(b) Begin and end with an R?
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document.write( "These are the arrangements of two letters from FOEVE with an R added \r\n" );
document.write( "on each end.\r\n" );
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document.write( "Case 1: use both E's between two R's, i.e., REER\r\n" );
document.write( "That 1 way for case 1\r\n" );
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document.write( "Case 2: use 2 letters from FOEV in between the 2 R's.\r\n" );
document.write( "That's \"4P2+=+4%2A3+=+12\"\r\n" );
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document.write( "Total for both cases: 1+12 = 13 ways\r\n" );
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(c) Had the letters EVER?
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document.write( "If all the letters of EVER were distinguishable it would 4!, but since there\r\n" );
document.write( "are 2 indistinguishable E's, we divide by 2!, so the answer is\r\n" );
document.write( "\"4%21%2F2%21+=+24%2F2+=+12\"\r\n" );
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(d) did not have any repeated letters?
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document.write( "That's Case 1 of part (a) \r\n" );
document.write( "Answer: 120

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document.write( "Case 1:  Both R's between vowels:  ORRE, ERRO, ERRE \r\n" );
document.write( "That's 3 for Case 1.\r\n" );
document.write( "Case 2:  E--E with two different letters between the E's taken\r\n" );
document.write( "from FORV.\r\n" );
document.write( "That's \"4P2+=+4%2A3+=+12\"\r\n" );
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document.write( "Case 3:  E--O with two different letters between the E's taken\r\n" );
document.write( "from FREV,\r\n" );
document.write( "That's also 12 \r\n" );
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document.write( "Case 4:  O--E with two different letters between the E's taken\r\n" );
document.write( "from FREV,\r\n" );
document.write( "That's also 12\r\n" );
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document.write( "Total from all four cases: 3+12+12+12 = 39 ways.\r\n" );
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document.write( "Edwin
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