Permutations And Combinations Question 250

Question: Find the total number of 9 digit numbers which have all the digits different [IIT 1982]

Options:

A) $ 9\times 9\ ! $

B) $ 9\ ! $

C) 10!

D) None of these

Show Answer

Answer:

Correct Answer: A

Solution:

  • There are 10 digits in all viz. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The required 9 digit numbers = (Total number of 9 digit numbers including those numbers which have 0 at the first place) - (Total number of those 9 digit numbers which have 0 at the first place) $ ={{}^{10}}P_9-{{}^{9}}P_8=\frac{10!}{1!}-\frac{9!}{1!}=10!-9! $ = $ (10-1)9!=9.9! $ .

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