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7!/(3!*2!) = 7*6*5*4*3*2*1 / 3*2*1*2*1 which is equal to 420
\n" ); document.write( "7! if all the letters were different.
\n" ); document.write( "Since 3 O's are common, you have to divide by 3!.
\n" ); document.write( "Since 2 T's are common, you have to divide by 2!.
\n" ); document.write( "Hard to see with 7 letters.
\n" ); document.write( "Easier to see with 3.
\n" ); document.write( "Assume letters are a,b,c
\n" ); document.write( "possible permutations are:
\n" ); document.write( "a,b,c
\n" ); document.write( "a,c,b
\n" ); document.write( "b,a,c
\n" ); document.write( "b,c,a
\n" ); document.write( "c,a,b
\n" ); document.write( "c,b,a
\n" ); document.write( "that's a total of 6 which is equal to 3!
\n" ); document.write( "Now assume 2 of the letters are the same.
\n" ); document.write( "let's assume it's b.
\n" ); document.write( "then the letters are a,b,b
\n" ); document.write( "possible permutations are:
\n" ); document.write( "a,b,b
\n" ); document.write( "b,a,b
\n" ); document.write( "b,b,a
\n" ); document.write( "that's a total of 3 which is equal to 6! / 2! = 3
\n" ); document.write( "same concept applies to 7 letters but the number of permutations is too large to list separately.
\n" ); document.write( "TORONTO has 2 of the letter T and 3 of the letter O and 1 each of the rest.
\n" ); document.write( "formula becomes 7! / (3!*2!) which is equal to 7*6*5*4*3*2*1 / 3*2*1*2*1 which is equal to 7*6*5*2 which is equal to 420.
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