Permutation of Indistinguishable Objects How many number of two digit numbers can be formed using $\{4,5,6,6\}$ without repetition? I know that $\{45,46,54,56,65,64,66\}$ are the possible answers, but I am wondering if there is any formula that can be used to get this. I tried this formula $\dfrac{4P2}{2!}$which I found here. 
 A: One way to think about it is to consider $66$ as only being one $6$, giving you $3P2 = 6$ cases, and then add the extra case of having the number $66$. You can do this because you know all other permutations resulting from the additional $6$ will be repeats. 
A: Imagine you had four distinguishable digits, $a$, $b$, $c$ and $d$.  The number of two-digit numbers you could create from these distinguishable digits is clearly $4 \times 3 = 12$---you could choose any of the four for the first digit, and any of the remaining three as the second digit.
But for your case, in which the last two available digits ($c$ and $d$) are indistinguishable, the previous answer ($12$) represents an overcounting.  Why?  For any case that had a single $c$ there is an equivalent case that contains a $d$ in the given position; you should only count half of those cases as distinguishable.  How many such "ambiguous" cases are there?  Of those putative $12$, only $2$ cases have no $c$ or $d$ (the number $ab$ and $ba$)---the rest have a $c$ or $d$.  Thus there are $10$ cases with a single $c$ or a single $d$.  Where you had a $c$ you also have an equivalent case where you have a $d$.  So of these $10$ cases that contain a single $c$ or a single $d$, half are numerically equivalent.  In other words, you have overcounted by half of 10.  Thus the final answer is $12 - 10/2 = 7$.
