Answer: Option B
:
B
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.