document.write( "Question 1160524: HOW MANY DIFFERENT 10-LETTER WORDS(REAL OR IMAGINARY) CAN BE FORMED FROM THE LETTERS IN THE WORD REPETITION \n" ); document.write( "

Algebra.Com's Answer #783827 by jim_thompson5910(35256) $\"\"$ $\"About$

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\n" ); document.write( "For the word \"repetition\", we have 10 letters. So there are
\n" ); document.write( "10! = 10*9*8*7*6*5*4*3*2*1 = 3,628,800
\n" ); document.write( "different ways to arrange them, but only if we can tell the two 'e's apart, and the same goes for the 'i's and the 't's as well. \r
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\n" ); document.write( "\n" ); document.write( "Since we can't distinguish these letters, we have to divide by 2! = 2*1 = 2 for each repeated letter. This is to avoid double counting per either the 'e's, 'i's or 't's.\r
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\n" ); document.write( "\n" ); document.write( "So we have (3,628,800)/(2!*2!*2!) = (3,628,800)/(2*2*2) = 453,600 permutations when we cannot distinguish between the repeated letters mentioned above. \r
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\n" ); document.write( "\n" ); document.write( "Final Answer = 453,600\r
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