JEE Main 2024 — 9 April, Shift 2 — Question 24

We are to find the number of integers between $100$ and $1000$ whose sum of digits is $14$.

Step 1: Represent the number

Let the three-digit number be $\overline{abc}$, where

$$1 \le a \le 9, \quad 0 \le b \le 9, \quad 0 \le c \le 9$$

and

$$a+b+c=14.$$

Put $a' = a-1$. Then

$$a'+b+c=13, \quad a', b, c \ge 0.$$

Step 2: Count non-negative integer solutions

The number of non-negative integer solutions of

$$a'+b+c=13$$

is

$$\binom{13+3-1}{3-1} = \binom{15}{2} = 105.$$

Step 3: Subtract invalid cases

(i) When $a \ge 10$ ($a' \ge 9$)

$$a''+b+c=4$$ $$\Rightarrow \binom{6}{2}=15.$$

{(ii) When $b \ge 10$

$$a'+(b-10)+c=3$$ $$\Rightarrow \binom{5}{2}=10.$$

(iii) When $c \ge 10$

$$a'+b+(c-10)=3$$ $$\Rightarrow \binom{5}{2}=10.$$

No two cases can occur simultaneously.

Total invalid cases:

$$15+10+10=35.$$

Step 4: Required count

$$105-35=70.$$ $$\boxed{70}$$