Permutations and Combinations – Number Theory – JEE Main 28 Jan 2025 Shift 2

Solution:

Let the number be $N = xyz$ where $x, y, z$ are digits.
The range is $212 \le N \le 999$.
The sum of digits $x + y + z = 15$.

Since $x$ is the hundreds digit, $x \in \{2, 3, \dots, 9\}$.
We analyze case by case for $x$:

1. **If x = 2**:
$y + z = 13$. Since $y, z \le 9$, possible pairs $(y, z)$ are:
$(4, 9), (5, 8), (6, 7), (7, 6), (8, 5), (9, 4)$.
Total = 6. (All numbers $>212$).

2. **If x = 3**:
$y + z = 12$. Pairs: $(3, 9)$ to $(9, 3)$.
Total = 7.

3. **If x = 4**:
$y + z = 11$. Pairs: $(2, 9)$ to $(9, 2)$.
Total = 8.

4. **If x = 5**:
$y + z = 10$. Pairs: $(1, 9)$ to $(9, 1)$.
Total = 9.

5. **If x = 6**:
$y + z = 9$. Pairs: $(0, 9)$ to $(9, 0)$.
Total = 10.

6. **If x = 7**:
$y + z = 8$. Pairs: $(0, 8)$ to $(8, 0)$.
Total = 9.

7. **If x = 8**:
$y + z = 7$. Pairs: $(0, 7)$ to $(7, 0)$.
Total = 8.

8. **If x = 9**:
$y + z = 6$. Pairs: $(0, 6)$ to $(6, 0)$.
Total = 7.

Total number of such numbers:
$$6 + 7 + 8 + 9 + 10 + 9 + 8 + 7 = 64$$

Ans. (64)