Permutations with repetition leading to fraction answer I have the following scenario where I need to find the number of unique 4 digit numbers using the following 6 digits:
4,4,4,4,6,6
According to my understanding this can be derived by the following calculation:
6P4 / 4! . 2!
However this give the following answer 7.5?
Any help will be much appreciated.
Thanks
 A: This corresponds to counting the number of 4-letter words formed from AAAABB. We have three cases:


*

*AAAA which constitutes a single solution

*AAAB and permutations ${}_4 C_1=4$

*AABB and permutations ${}_4 C_2=6$
This gives us a total of $11$ words.

There is a problem with the model in the link you gave:
$$
\frac{{}_n P_r}{a!b!\cdots}
$$
where $a,b,...$ are the number of repetitions of the different items.
The problem is that this model breaks down whenever the number of unchosen items $n-r$ is greater than $2$ while there is at least one repeated item in the list. The reason for this is that the figures in the denominator also count permutations of the $n-r$ items that where NOT chosen, but the numerator only counts permutations of the $r$ chosen items.

This is why the figure:
  $$
\frac{{}_6 P_4}{4!2!}=7.5
$$
  that you gave is too low.
  
  
*
  
*The numerator ${}_6 P_4$ counts $4444$ as $4!$ permutations.
  
*Whereas the denominator counts $4444$ not choosing $66$ as $4!2!$ permutations.
  
*So instead of contributing one arrangement, the case $4444$ contributes only $1/2$ arrangement.
  

Similarly, the numerator counts $4466$ as ${}_4 P_2 2!=4!$ permutations, but divides them by $4!2!$ counting only half an arrangement. The same can be said for each of the ${}_4 P_4/(2!2!)=6$ rearrangements of $4466$. So those $6$ cases are only counted as half as many, namely $3$.

Conclusion: The method given in the link is useless when the difference between $n$ and $r$ is greater than $1$.
