Question

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

Answer 1

After fixing the places of three pearls. Treating $3$ specific pearls $=1$ units, 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 anticlockwise arrangements.

So the required number of arrangements $=\frac{1}{2}18!.6!=3.18!$