Integers Between 100 and 1000 With Digit Sum Equal to 14

JEE Main 2024 — 9 April, Evening Shift. Previous Year Question.

Problem

The number of integers between $100$ and $1000$ having the sum of their digits equal to $14$ is ____.

Key insight. A 3-digit number has three digits, each within its own range — the hundreds digit can’t be $0$, but the other two can. Counting solutions to $a+b+c=14$ without those range limits overcounts; the fix is inclusion-exclusion, subtracting off exactly the cases where a digit is pushed past $9$.

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Approach

Let the number be $\overline{abc}$ with $a \in \{1,\dots,9\}$ and $b,c \in \{0,\dots,9\}$, and $a+b+c=14$. Shift $a$ so every variable starts from $0$, count all non-negative integer solutions using stars-and-bars, then subtract off the solutions where any one digit exceeds $9$ using inclusion-exclusion.

Solution

Step 1 — Set up the equation and shift the hundreds digit

Let $a’=a-1$, so $a’ \in \{0,\dots,8\}$. The equation $a+b+c=14$ becomes:

$$a’+b+c = 13, \qquad 0 \le a’ \le 8,\ 0 \le b,c \le 9$$

Step 2 — Count all non-negative solutions, ignoring upper bounds

By stars-and-bars, the number of non-negative integer solutions to $a’+b+c=13$ is:

$$\binom{13+2}{2} = \binom{15}{2} = 105$$

Step 3 — Subtract the cases where a variable exceeds its bound

$a’ \geq 9$: substitute $a”=a’-9$, giving $a”+b+c=4$, with $\dbinom{4+2}{2}=\dbinom{6}{2}=15$ solutions.

$b \geq 10$: substitute $b’=b-10$, giving $a’+b’+c=3$, with $\dbinom{3+2}{2}=\dbinom{5}{2}=10$ solutions.

$c \geq 10$: by symmetry with $b$, also $10$ solutions.

Step 4 — Check for double-counted overlaps

Could two variables exceed their bounds simultaneously? For instance, $a’ \geq 9$ and $b \geq 10$ together would need $a”+b’+c = 13-9-10 = -6$, which is impossible. Every other pairwise overlap is similarly impossible, since two excesses together already exceed $13$. So nothing needs to be added back.

Step 5 — Apply inclusion-exclusion

$$\text{Valid solutions} = 105 – 15 – 10 – 10 = 70$$

Answer

Common mistakes

• Forgetting that $a$ (the hundreds digit) starts from $1$, not $0$. Treating all three digits as ranging over $0$–$9$ without the shift changes the equation’s constant and throws off every subsequent count.
• Skipping the overlap check in inclusion-exclusion. Even when the overlap turns out to be zero, verifying that two excess conditions can’t hold simultaneously is what makes the subtraction step valid rather than a lucky guess.

Practise next

• Find the number of integers between $1000$ and $10000$ whose digits sum to $20$, using the same shifted stars-and-bars approach with a fourth digit.