Permutations And Combinations Question 46

Question: How many numbers between 5000 and 10,000 can be formed using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 each digit appearing not more than once in each number [Karnataka CET 1993]

Options:

A) $ 5{{\times }^{8}}P_3 $

B) $ 5{{\times }^{8}}C_3 $

C) $ 5\ !\ {{\times }^{8}}P_3 $

D) $ 5\ !\ {{\times }^{8}}C_3 $

Show Answer

Answer:

Correct Answer: A

Solution:

  • A number between 5000 and 10,000 can have any of the digits 5, 6, 7, 8, 9 at thousand’s place. So thousand’s place can be filled in 5 ways. Remaining 3 places can be filled by the remaining 8 digits in $ ^{8}P_3 $ ways. Hence required number = $ 5{{\times }^{8}}P_3 $ .