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There are 2 brothers among a group of 20 persons. In how many ways can the group be arranged around a circle so that there is exactly one person between the two brothers?


2 × 17!


18! × 18


19! × 18


2 × 18!


2 × 17! × 17!

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Option(D) is correct

Circular Permutation

$n$ objects can be arranged around a circle in $(n - 1)!$

If arranging these $n$ objects clockwise or counter clockwise means one and the same, then the number arrangements will be half that number.

i.e., number of arrangements $= \dfrac{(n-1)!}{2}$ 

Let there be exactly one person between the two brothers as stated in the question.

If we consider the two brothers and the person in between the brothers as a block, then there will 17 others and this block of three people to be arranged around a circle.

The number of ways of arranging 18 objects around a circle is in $17!$ ways. 

Now the brothers can be arranged on either side of the person who is in between the brothers in $2!$ ways.

The person who sits in between the two brothers could be any of the 18 in the group and can be selected in 18 ways.

Therefore, the total number of ways,

$18 × 17! × 2$

$=\textbf{2 × 18!}$

(1) Comment(s)


Can you explain it again with every steps? Please