Question: The number of ways in which ten candidates $ A_1,\ A_2,\ …….A _{10} $ can be ranked such that $ A_1 $ is always above $ A _{10} $ is
Options:
A) $ 5\ ! $
B) $ 2(5\ !) $
C) $ 10\ ! $
D) $ \frac{1}{2}(10\ !) $
Show Answer
Answer:
Correct Answer: D
Solution:
- Without any restriction the 10 persons can be ranked among themselves in $ 10\ ! $ ways; but the number of ways in which $ A_1 $ is above $ A _{10} $ and the number of ways in which $ A _{10} $ is above $ A_1 $ make up $ 10\ ! $ . Also the number of ways in which $ A_1 $ is above $ A _{10} $ is exactly same as the number of ways in which $ A _{10} $ is above $ A_1 $ . Therefore the required number of ways $ =\frac{1}{2}(10\ !) $ .
BETA
