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So using numbers from 0~9, making a 5 digit number, how many numbers can be formed that is bigger than 12345?

Repetition is not allowed.

Thank you.

Sadpersonn
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    What have you tried so far? You may want to try thinking about how you can pick on digit at a time: if you pick say $2$ as the first digit, how many numbers can you create? If you pick $1$ as the first, what can you pick for the second? – Brian61354270 Apr 27 '19 at 23:39
  • I know that If I pick 1 as the first, I can pick 2~9 for the second, but it doesn't mean that I can pick 3~9 for the third. This is because If I picked 3 for the second, I can pick 0 for the third. (for ex 13082). – Sadpersonn Apr 27 '19 at 23:53
  • 9x9x8x7x6= 27216 and 1x2x2x2x2= 16, so 27216-16=27200. Is this correct? – Sadpersonn Apr 28 '19 at 00:02

2 Answers2

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Finding the number of $5$ digit numbers without repetition and how many of said numbers $\leq 12345$ would gives us the number of numbers without repetition $> 12345$.

Number of $5$ digit numbers without repetition = $9\cdot9\cdot8\cdot7\cdot6 = 27216$.

The number of $5$ digit numbers without repetition $\leq12345$ has 4 cases:

  1. The number of $5$ digit numbers without repetition of the form $10$_ _ _ $= 8\cdot7\cdot6$
  2. The number of $5$ digit numbers without repetition of the form $120$ _ _$=7\cdot6$
  3. The number of $5$ digit numbers without repetition of the form $1230$ _$=6$
  4. The number of $5$ digit numbers without repetition of the form $1234$ _$=2$

Number of $5$ digit numbers without repetition greater than $12345= 27216-8\cdot7\cdot6-7\cdot6-6-2=26830$

Darius
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Or, to do it directly, we have $8\cdot 9\cdot 8\cdot7\cdot6+7\cdot 8\cdot 7\cdot 6+6\cdot 7\cdot 6+5\cdot 6+4=24192+2352+252+30+4=26830$.

Here I have just added the number of five digit numbers without repitition greater than $12345$ of the form $abcde,\, a\gt1$, $1bcde,\,b\gt2$, $12cde,\,c\gt3$, $123de,\,d\gt4$ and, finally, $1234e,\,e\gt5$.