Question

Find the total number of 9 digit numbers which have all the digits different.

• A. $9 \times 9\mspace{6mu}!$
• B. $9\mspace{6mu}!$
• C. 10 !
• D. None of these

Answer 1

Answer: (A)

$9 \times 9\mspace{6mu}!$

Answer 2

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)

$=^{10}P_{9} -^{9}P_{8} = \frac{10!}{1!} - \frac{9!}{1!} = 10! - 9!$ = $(10 - 1)9! = 9.9!$.