Question

Consider 21 different pearls on a necklace. How many ways the pearls can be placed in this necklace such that three specific pearls always remain together?
A.18!
B.3(18!)
C.21!
D.None of these

Answer 1

Hint: There are 21 pearls but 3 specific pearls are always together. Thus these three pearls will make a group. Thus total pearls will become 18. Using permutations or combinations will help us in finding the different ways of arrangements.

Complete step-by-step answer:
Given that there are 21 pearls in the necklace.
And there are three pearls that need to be remained together always.
So number of patterns = \[\left( {21 - 3} \right)! = 19!\]
But we need to treat them as one special unit.
Now we have \[\left( {19 - 1} \right)! = 18!\] ways.
But talking about those three pearls they also can be arranged in different ways among themselves.
So their arrangement will be, \[3! = 6\] ways.
Now arrangement can be either clockwise or anticlockwise. But it is not mentioned in the problem.
So looking on overall factors, the number of ways in which the pearls can be arranged is given by
\[
   \Rightarrow \dfrac{1}{2} \times 6 \times 18! \\
   \Rightarrow 3 \times 18! \\
\]
So option B is the correct answer.

Note:Here 3 specific pearls are always together. Consider that as one. Don’t directly jump to conclusions.
Consider all conditions when you solve a problem related to combinations and arrangements.