## 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’.