We need a number of the form ABC where $A+B+C\le 5$, and A,B,C are whole numbers.

Method 1 - Trial and Error - Manual approach

List out all the possibilities manually,

Case 1: When sum of digits=1

A+B+C=1

We can solve this using the similar to different approach where there is 1 zero and 1 one. Number of combinations ${=}^3{C}_2=3$

Case 2: When sum of digits=2

A+B+C=2

Number of possibilities = Answer based on 2 zeroes and 2 ones ${=}^4{C}_2=6$

Case 3: When sum of digits=3

A+B+C=3

Number of possibilities = Answer based on 3 zeroes and 2 ones ${=}^5{C}_2=10$

Case 4: When sum of digits=4

A+B+C=4

Number of possibilities = Answer based on 4 zeroes and 2 ones ${=}^6{C}_2=15$

Case 5: When sum of digits=5

A+B+C=5

Number of possibilities = Answer based on 5 zeroes and 2 ones ${=}^7{C}_2=21$

Total = 21+15+10+6+3 = 55

Method 2- Shortcut

We will use a dummy variable "D” to take the place of the $\le$ sign.

Then the equation becomes -

A+B+C+D=5 (Where D can take values 0,1,2,3,4,5)
$\begin{array}{cc}A+B+C& D\\ 5& 0\\ 4& 1\\ 3& 2\\ 2& 3\\ 1& 4\\ 0& 5\end{array}$

Using the same Similar to Different approach, we can get the answer based on 5 zeroes and 3 ones as ${}^8{C}_3=56$

We must remove the case where D=5, because in that case A+B+C=0, which is not possible. Thus, the answer is 56-1=55.