Permutation of $n$ objects taken $r$ at a time having $p$ similar objects What is the permutation of $n$ objects taken $r$ at a time having $p$ similar objects?
Consider the example
What is the number of ways that $5$ red balls and $2$ black ball can be arranged taken $2$ at a time?
If you consider the probable combination it will be nothing but
$\{RB,BR,RR,BB\}$ and nothing other else. So how can we get $4$ ways by formula.
As we know $P$ of $n$ DISTINCT object taken $r$ at a time is given by
$$^n P_r =  \frac{n!}{(n-r)!}$$
And permutations of $n$ objects having $p$ similar objects and taken $r$ object at a time is given by
$$^n P_r = \frac{n!}{(n-r)! \cdot p!}$$
Here if we put that, it comes out as
$$\frac{7\times6}{5! \times 2!}$$
But it's a fraction:
$$\frac{42}{120 \times 2} = \frac{21}{120}$$
and not 4 
How where am I wrong.
 A: Chosing subsets its not about permutations and repeated permutations, but it's just about combinations. Your problem is quite easy to understand practically - answer: $3$ ($\{B, R\}, \{R, R\}, \{B, B\}$). Remember that (unordered) sets chosen don't count distinctly if you change order ($\{B, R\} = \{R, B\}$).
To solve this, in general case, with $k$ type of objects, $m$ objects in a subset, $n_m$ objects with at least $m$ occurencies, $n_{m - 1}$ objects with at least $m - 1$ occurencies and so on may not be as easy as understanding a practical idea, as these require analytical solutions (there are combinatorial objects regarding multisets, not simple sets).
Simple is to understand and solve the case with  $\alpha$ types of objects with at least $2$ occurences and $\beta$ types of objects with $1$ occurence, chosen in . There are $\binom{\alpha}{2}$ ways to choose pairs of identical objects and $\binom{\alpha + \beta}{2}$ ways to choose pairs of distinct objects. So, our combinatorial object is defined as $\binom{\alpha, \beta}{2} = \binom{\alpha}{2} + \binom{\alpha + \beta}{2}$, a generalisation of standard combinations.
Note that $\binom{n}{r}$ is the binomial coefficient, an expression of combinations. You should read more on these topics to understand the difference between different combinatorial objects like permutations, repetition permutations, combinations, arrangeaments (here order is important!) and nultiset combinatorics.
A: I think you answered your own question, isn't it? You stated "As we know P of n DISTINCT object taken r at a time is given by..." So then, in your case, there are 2 DISTINCT objects, namely, red balls and black balls, taken 2 at a time. Hence, nPr would be 2!/(2-2)! + 2!/0! = 2! = 4.
