How many number less than $100000$ contain digits ${3,4,5}$ such that each number contains each digit of ${3,4,5}$ once time? 
How many number less than $100000$ contain digits  ${3,4,5}$ such that each number contains each digit of ${3,4,5}$ once time ? 

My try follows: -
There are 3 cases 
First : 5 digits numbers: $5P3×10×10$
Second:  4 digits numbers : $4P3×10$
Third: 3 digits numbers:  there are only $6$ 3-digits 
Is my work  true ? 
Thank you for your help 
 A: You can uniquely define such a number by chosin the 3 among 5 digits that are 3, 4, 5, you then have 7 choices for both other digits, and 6 choices for the permutations of your three digits. Thus you get :
$${5\choose 3}\times 7^2\times 6 = 2940$$
In fact you do not need to make the distinction between the cases as you did, since a 3 digit number can be considered as a 5 digit number by adding 0s at the beginning.
Also, you should not have multiplied by 10, but by 7, since once you have chosen the digits that are 3, 4, 5, the other digits can only be 0, 1, 2, 6, 7, 8, 9, ie you only have 7 choices.
Finally, as you did in you last step, you needed to multiply each result by six (and not only the last, ie $3P3 \times 6 = 6$).
However making your method work can be tedious since you need to make sure that the first digit of you number is not zero (while the others can be).
A: Firstly we choose the 3 places for 5,4,3 from the 5 places. There are 5C3 ways. Then we permude them in the 3 places,i.e. simply we get 5P3 ways of having 5,4,3 in the required numbers. In the remaining 2 plces we can only place the digits excluding 5,4,3( 7 digits including 0). Thus total number of reqired nubers are 5P3 * 7*7 . 
This includes all cases, as if we don't choose the 1st digit and get 0. Then we get all such 4 digit numbers and similarly the 3 digit number.
