Question

The number of ways in which $20$ different pearls of two colours can be set alternately on a necklace, there being $10$ pearls of each colour, is

• A. $9!\times 10!$
• B. $5(9!{)}^2$
• C. $(9!{)}^2$
• D. none of these

Answer 1

Answer: (B)

$5(9!{)}^2$

Let's first place the pearls of one color in the necklace.
This can be done in $\frac{10!}{10}=9!$ ways.
(rule of circular probability)
Now just permute the 10 remaining pearls of the other kind in the 10 gaps created by the placement of pearls earlier such that different coloured pearls are alternate.
This can be done in $10!$ ways.
But since such arrangement in necklace is symmetrical we will have to divide by $2$
(otherwise same thing will get repeated twice because of symmetry)
Hence total number of ways=$\frac{9!\times 10!}{2}=\frac{10\times \left({9!}^2\right)}{2}=5\times \left({9!}^2\right)$
Hence, option 'B' is correct.