Question

In a set of n things ${}^{\prime }{r}^{\prime }$ things are similar and remaining are different. Then the number of circular arrangements of those $n$ things is

• A. $(n-1)!r!$
• B. $\frac{(n-1)!}{r!}$
• C. $\frac{(n-1)!}{r}$
• D. $r(n-1)!$

Answer 1

Answer: (B)

$\frac{(n-1)!}{r!}$

$\Rightarrow$Circular arrangement of $n$ things is $(n-1)!$

And out of $n$ things $r$ things are similar

$\therefore$ Total arrangement is $\frac{(n-1)!}{r!}$

Hence option ${}^{\prime }{B}^{\prime }$ is the answer.