Permutation of $n$ objects taken $r$ at a time having $p$ similar objects

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

Let us name the 5 red balls as R1, R2, R3, R4, R5 and the 2 black balls as B1, B2. Now, let us, temporarily, treat R1, R2, R3, R4, R5 as different balls and B1, B2 as different balls. Then, the number of permutations of 7(=n) different balls taken 2(=r) at a time, and the balls do not repeat is given by nPr=n!/(n−r)! 7P2= 7!/5!= 7x6= 42. But we know that the balls are not all different.Thus, let us now treat R1, R2, R3, R4, R5 as the same red balls and B1, B2 as the same black balls. Then, the 20 arrangements {R1,R2}, {R1,R3}, {R1,R4}, {R1,R5}, {R2,R1}, ... , {R2,R5}, {R3,R1}, ... , {R3,R5}, {R4,R1}, ... , {R4,R5}, {R5,R1}, ... , {R5,R4} would be exactly the same as one another and can be written as one of four permutations {R,R}. Similarly, the 2 arrangements (B1,B2}, {B2,B1} would be exactly the same as one another and can be written as one of four permutations {B,B}. Now, each of the remaining two permutations {R,B} and {B,R} would have 10 corresponding arrangements ( do them for yourself). Thus ,we can account: 20+2+10+10=42 arrangements ( if we treat the balls as different) have been condensed into 4 permutations.

Now, let us see where you committed a mistake: nPr=n!/(n−r)!⋅p! There is indeed no such formula for the number of permutations of n objects( in which p1 are of one kind, p2 are of second kind... pk are of kth kind and the rest, if any, are different) for r objects taken at a time. Yes, if all the n objects are taken at a time, then we have Required number of permutations= n!/p1!xp2!x...xpk!

Thank you,