Question

A bangle is studded with 18 pearls of different colours. What is the total number of ways in which pearls studded in the bangle can be arranged so that there is always one pearl between a red and a blue pearl?

Answer 1

Answer:

Possible number of ways of arranging the pearls in given condition is 2 × 16!  ways .

Step-by-step explanation:

Number of ways in which red and blue pearl are arranged = 2 ways

Number of ways in which other pearls can be arranged between red and blue = 16 ways

Remaining 15 pearls can be arranged in 15! Ways

Hence, possible number of ways of arranging the pearls = 2 × 16 × 15! = 2 × 16! = $4.1845579776\times10^{13}$ ways