SATHEE: Permutations And Combinations Question 290

Permutations And Combinations Question 290

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\ !) $ .

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Permutations And Combinations Question 290

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:

Answer:

Solution:

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