Question

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

• A. $18!$
• B. $3(18!)$
• C. $21!$
• D. None of these

Answer 1

The correct option is B $3(18!)$

After fixing the places of three pearls.

Treating $3$ specific pearls = $1unit$, so we have now $18pearls+1unit=19$ 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 anticlockwise arrangements.
So, the required number of arrangements

$=\frac{1}{2}18!.6=3(18!)$.