How many distinct permutations can be formed using the letters of the word tallahassee?

831600



Tallahassee has 11 letters therefore there are 11! permutations that can be made.

11! = 11x10x9x8x7x6x5x4x3x2x1 = 39916800 permutations



This is because the first letter can be any of 11

The second letter can be any of 10, one is already in place

The third can be any od 9 the first two are already in place

The 4th can be any of 8 the first three are already in place

etc etc



Multiply them all together to give the number of permutations.

However the question doesn't ask for the number of permutations but instead the number of "distinct permutations".



Eg for explanation purposes i shall label the two "S"s S1 and S2.

One permutation of the letters would be TALLAHA"S1""S2"EE but this cannot be differentiated in practice from TALLAHA"S2""S1"EE. The letters are all in the same positions just that the two 's"s have swapped places. However, in the above calculation these are treated as separate permutations. These may be "separate" but they are not "distinct". Therefore we have to divide 11! to allow for these occurances. Please note for 50% of the 11! permutations there is an exact copy with only the 2 "S"s changing position. Therefore we have to divide 11! by 2 to allow for the two "S"s.



There are also 2 x "E"s. Therefore we must also divide by 2 to allow for the multiple "E"s in the same way as the "S"s.

There are 2 "L"s in addition - therefore divide by a further 2 to allow for this fact.



There are three "A"s. Here I shall call the different "A"s 1,2 and 3.

The As can be placed in 6 different ways for each combination of the other letters.

ie 123 = 132 = 231 = 213 = 312 = 321

The first of the 3As can be anyof 3, the second any of 2 and the last no choice 3x2x1 = 6 or 3!

Therefore we need to divide the original 11! by 3! to allow for the multiple "A"s.



(Please note when we are dividing by 2 for the Es and Ss and Ls we are really dividing by 2!, the first can be either of 2, the second no choice, 2x1 = 2 = 2!)



Therefore there are 11! / 2!2!2!3! "distinct" permutations.

= 39916800 / 2x2x2x3x2 = 39916800/48 = 831600How many distinct permutations can be formed using the letters of the word tallahassee?
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