document.write( "Question 949395: How many different letter arrangments can be made from the leters in the word
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Algebra.Com's Answer #579613 by Edwin McCravy(20055) $\"\"$ $\"About$

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document.write( "PROBABILITY\r\n" );
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document.write( "There are 11 letters in PROBABILITY.\r\n" );
document.write( "if it were spelled\r\n" );
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document.write( "PROBAbIliTY, then the answer would be 11!.  For spelled\r\n" );
document.write( "that way it has one of the B's capital and the other b \r\n" );
document.write( "a small letter (the same for the I and i). Then we could \r\n" );
document.write( "tell the difference between, say, these 4 arrangements:\r\n" );
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document.write( "iOYIbRAPBTL, iOYIBRAPbTL, IOYibRAPBTL, and IOYiBRAPbTL, \r\n" );
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document.write( "However, since the 2 B's are indistinguishable, as well \r\n" );
document.write( "as the 2 I's, we cannot tell those 4 arrangements apart.\r\n" );
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document.write( "All 4 look like IOYIBRAPBTL.\r\n" );
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document.write( "So we must divide 11! by 4.  That is, we divide 11! by the \r\n" );
document.write( "factorials of the numbers of times each letter occurs \r\n" );
document.write( "in the given word.  \r\n" );
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document.write( "Since B and I each occur 2 times in PROBABILITY, we divide \r\n" );
document.write( "by 2! twice, once for B, and once for I, since each occurs \r\n" );
document.write( "2 times, So we divide by 2! twice which amounts to dividing \r\n" );
document.write( "by 2!2! =2*2 or 4. That's why we divide by 4.  So the \r\n" );
document.write( "answer is\r\n" );
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document.write( "Edwin

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