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# In a set of n things 'r' things are similar and remaining are different. Then the number of circular arrangements of those n things is\begin{align} & A.\left( n-1 \right)!r! \\ & B.\dfrac{\left( n-1 \right)!}{r!} \\ & C.\dfrac{\left( n-1 \right)!}{r} \\ & D.r\left( n-1 \right)! \\ \end{align}

Last updated date: 18th Jun 2024
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Hint: In this question, we need to find a number of ways n things can be arranged such that r things are similar and remaining are different. For this, we will use the derivative of permutation. We will use the following derivations.
(1) The total number of circular permutations of n different things taken all at a time is (n-1)!.
(2) If any item is repeated then we divide the total arrangements by that repeated item.

Complete step-by-step solution
Here, we have to find the number of circular arrangements of n things out of which r things are similar and remaining is different.
We know that the total number of circular permutations of n different things taken all at a time is (n-1)!.
So the total number of circular arrangements will be (n-1)!.
Now out of these n things, r things are similar, so we need to divide total arrangements by repeated things.
Since r things are similar, so the number of arrangements of r items will be r!
Now we know that, when we have repeated items, then we divide total arrangements by repeated items to get the final answer.
Here, total circular arrangements are (n-1)! and repeated arrangements are r!
So our final circular arrangements will be equal to $\dfrac{\left( n-1 \right)!}{r!}$.
Hence, $\dfrac{\left( n-1 \right)!}{r!}$ is our final answer.
Therefore, option B is the correct answer.

Note: Students can make the mistake of taking permutation repeated items as “r” but we need to take permutation as r! Make sure to divide the repeated thing. If we were given r things of one type repeated and p things of another type repeated then, we will divide total arrangements by $p!\times r!$ Hence total circular arrangements would be $\dfrac{n!}{p!r!}$.