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

A 9! × 10!

B 5(9!)²

C (9!)²

D (18!)²

E None of these

Correct Answer: Option B

First of all we will make circular permutations for all the flowers of a particular colour.

We know that, for permutation of n objects in a circle is = (n - 1)!

So, for 10 flowers of same colour, the permutations in 10 places = (10 - 1)! = 9!

Now, for permutation of the remaining ten flowers of same colour in between the previously placed flowers = 10!

Now, since this is a necklace hence the number of permutations are divided by 2.

Hence, the required answer = (9! × 10!)/2 = 5(9!)²