Permutations And Combinations Question 351
Question: 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.
Options:
A) $ 6\times {{(9!)}^{2}} $
B) 12!
C) $ 4\times {{(8!)}^{2}} $
D) $ 5\times {{(9!)}^{2}} $
Correct Answer: D
Show Answer
Answer:
Solution:
$ \therefore $ Required number of ways $ =\frac{1}{2}\times 9!\times 10!=5{{(9!)}^{2}} $
BETA
