Consider $21$ different pearls on a necklace. How many ways can the pearls be placed in on this necklace such that $3$ specific pearls always remain together?

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$3 (1)$

$4 (18!)$

$3 (18)$

$3 (18!)$

answer is D.

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Detailed Solution

After fixing the places of three pearls, treating $3$ specific pearls =1 unit. So, we have now

$18$ pearls + $1$ unit = $19$ and the number of arrangement will be $(19 - 1)! = 18!$

Also, the number of ways of $3$ pearls can be arranged between themselves is $3! = 6 .$

Since, there is no distinction between the clockwise and anti-clockwise arrangements.

So, the required number of arrangements $= \frac{1}{2} 18! \cdot 6 = 3 (18!) .$

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