In how many different ways can you scramble the word ‘abracadabra’?

Apparently the solution is  11!  because ‘abracadabra’ has 11 letters. But this is not correct because replacing first ‘a’ with the last ‘a’ for example would not give a different solution, and by simply using permutations we count some words more than once.

The word ‘abracadabra’ contains following letters: ‘a’ – 5 times, ‘b’ – 2 times,  ‘r’ – 2 times,  ‘c’ – 1 time,  ‘d’ – 1 time.

To see the number of scrambles we can consider 11 empty positions which we fill with letters like this: 

1. First choose the positions where to place ‘a’ . There are: \binom {11} {5} =  \frac{11!}{6! \cdot 5!} distinct choices.

2. Then choose the positions where to place ‘b’. Here we have \binom {6} {2} =  \frac{6!}{2! \cdot 4!} choices because 5 from the total of 11 free position are already filled with ‘a’.

3. Choose the positions where to place ‘r’:  \binom {4} {2} =  \frac{4!}{2! \cdot 2!}

4. Choose the position for ‘c’:  \binom {2} {1} =  \frac{2!}{1! \cdot 1!}

5. The remaining position is for ‘d’:  \binom {1} {1} =  \frac{1!}{1! \cdot 0!}. This is of course 1 but later you’ll  see why I keep writing it as a formula.

Then the total is:

{\frac{11!}{6! \cdot 5!}} \cdot {\frac{6!}{2! \cdot 4!}} \cdot {\frac{4!}{2! \cdot 2!}} \cdot {\frac{2!}{1! \cdot 1!}} \cdot {\frac{1!}{1! \cdot 0!}} =  \frac{11!}{5! \cdot 2! \cdot 2! \cdot 1! \cdot 1!} = 83160

The blue formula above, shows actually a different way of looking at the problem. If we would count all the permutations (11!) then each permutation of ‘a’ letters will cause repeats. So we need to divide the total number of permutations to 5!, because we have 5 ‘a’ letters. From the same reason we need to divide with  2!  for ‘b’ letters, with  2!  for ‘r’ letters, and with  1!  for ‘c’ and ‘d’.

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