What are all the three-digit numbers that contain at least one of each digit?

 What are all the three-digit numbers that contain at least one of each digit?👇🏾👇🏾

To find all the three-digit numbers that contain at least one of each digit (0 through 9), we need to count how many such numbers exist.

First, note that a three-digit number can only have three distinct digits. Thus, we need to form numbers from the digits 0 through 9, ensuring that we use three different digits without repetition.

We have the following constraints:

1. The number cannot start with 0 because it must be a three-digit number.
2. The digits must be distinct.

To solve this, we can break it into the following steps:

1. Choosing three distinct digits: We need to select three distinct digits from the set $\{0,1,2,3,4,5,6,7,8,9\}$.

   • The number of ways to choose 3 distinct digits from 10 is given by the combination formula $\left(\genfrac{}{}{0px}{}{10}{3}\right)$.
   $$\left(\genfrac{}{}{0px}{}{10}{3}\right)=\frac{10!}{3!(10-3)!}=\frac{10\times 9\times 8}{3\times 2\times 1}=120$$

2. Arranging the chosen digits:

   • If 0 is one of the chosen digits, it cannot be in the hundreds place. Therefore, we need to count these separately.

3. Calculating the arrangements:

   • Case 1: 0 is not one of the digits:

     • If none of the digits is 0, then we have 9 choices for the hundreds place (1 through 9), and the other two digits can be arranged in the remaining two places.
     • The number of ways to arrange 3 digits where 0 is not included is $3!$ (since all positions are allowed for all three digits):
     $$3!=6$$
     • For choosing 3 digits from 9 (excluding 0), the number of combinations is $\left(\genfrac{}{}{0px}{}{9}{3}\right)$:
     $$\left(\genfrac{}{}{0px}{}{9}{3}\right)=\frac{9!}{3!(9-3)!}=\frac{9\times 8\times 7}{3\times 2\times 1}=84$$
     • Total arrangements in this case:
     $$84\times 6=504$$

   • Case 2: 0 is one of the digits:

     • We have to place 0 in one of the last two positions (tens or units place), not the hundreds place.
     • There are 2 positions for placing 0, and we have 2 other digits (from 1 to 9) to choose and place in the hundreds and the remaining place.
     • Choosing 2 digits from 9 non-zero digits:
     $$\left(\genfrac{}{}{0px}{}{9}{2}\right)=\frac{9!}{2!(9-2)!}=\frac{9\times 8}{2\times 1}=36$$
     • Each combination of digits (with 0) has $2!=2$ ways to arrange the two non-zero digits in the first two places:
     $$2!=2$$
     • Total arrangements in this case:
     $$36\times 2\times 2=144$$

4. Summing the results:

   • Total number of valid three-digit numbers is the sum of both cases:
   $$504+144=648$$

Thus, the total number of three-digit numbers that contain at least one of each digit is $648$.