A) \[6\times {{(9!)}^{2}}\]
B) 12!
C) \[4\times {{(8!)}^{2}}\]
D) \[5\times {{(9!)}^{2}}\]
Correct Answer: D
Solution :
[d] Ten pearls of one colour can be arranged in \[\frac{1}{2}.(10-1)!\] ways. The number of arrangements of 10 pearls of the other colour in 10 places between the pearls of the first colour =10! \[\therefore \] Required number of ways \[=\frac{1}{2}\times 9!\times 10!=5{{(9!)}^{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
