Question

The total 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. $6\times (9!{)}^2$
• B. $11!$
• C. $4\times (8!{)}^2$
• D. $5\times (9!{)}^2$

Answer 1

The correct option is D $5\times (9!{)}^2$

Ten pearls of one colour can be arranged in circular manner in $(10-1)!=9!$ ways.

The number of arrangements of $10$ pearls of other colour in $10$ places between the pearls of the first colour $=10!$

$\therefore$ Required number of ways $=\frac{9!\times 10!}{2}$
$(\because$ arrangement for anticlockwise and clockwise direction will not be considered different$)$
$=5\times (9!{)}^2$