Matematické Fórum

stuart clark · 24. 10. 2012 17:23

Find total numbers from $\bf{1}$ to $\bf{1000}$ in which all Digits are Distincts

JohnPeca18

You choose 3,2,1 numbers from 1-9 and permutates
$\sum_{i=1}^{3}\binom{9}{i}i!$
Now add numbers with 0. in 3-digit numbers yo can have 0 on 2 place, the rest you fill with 2 number from 1-9 and permutates. In 2 digit numbers you have 9 choices. So together it is
$\sum_{i=1}^{3}\binom{9}{i}i!+2*\binom{9}{2}*2+9$

vanok · 28. 10. 2012 14:18 — Editoval vanok (29. 10. 2012 11:49)

Hi ↑ stuart clark:,

Solution thanks to elementary combinatorial.

We have $9\cdot 9\cdot 8= 648$ numbers with 3 digits in which all digits are distinct.
We have $9\cdot 9=81$ numbers with 2 digits in which all digits are distinct.
We have $9$ numbers with 1 digit (in which all digits are distinct)
So, the answer to the question asked is $738=648+ 81+9$ numbers