Permutations and Combinations – Counting Principles – JEE Main 24 Jan 2026 Shift 1

Solution:

We need 4-digit numbers using digits $\{0, 1, 2, 5, 9\}$.
Range: $5000 < N < 9000$. First digit must be 5 (since it cannot be 9 as number is $<9000$, and cannot be 0, 1, 2). So, the number is of the form $5 \_ \_ \_$. Repetition is allowed. Sum of digits must be divisible by 3.
Let the digits be $5, a, b, c$.
Sum $S = 5 + a + b + c$.
We need $S \equiv 0 \pmod 3$.
$5 \equiv 2 \pmod 3$.
So we need $a+b+c \equiv 1 \pmod 3$.

Possible digits modulo 3:
$0 \to 0$
$1 \to 1$
$2 \to 2$
$5 \to 2$
$9 \to 0$
Let’s group digits by remainder:
$R_0 = \{0, 9\}$ (2 digits)
$R_1 = \{1\}$ (1 digit)
$R_2 = \{2, 5\}$ (2 digits)

We need $a+b+c \equiv 1 \pmod 3$.
Possible combinations of remainders for $(a, b, c)$:
1. $(1, 0, 0) \to 1+0+0=1$. Ways: $1 \times 2 \times 2 \times (\text{permutations of positions})$.
Number of choices for positions: The digits are chosen from sets $R_1, R_0, R_0$.
Choices for $(a,b,c)$:
– One from $R_1$, Two from $R_0$.
– Number of ways to choose digits sequence: $3 \text{ (positions for } R_1) \times 1 \text{ (choice from } R_1) \times 2 \text{ (choice 1 from } R_0) \times 2 \text{ (choice 2 from } R_0) = 12$.
2. $(1, 1, 2) \to 1+1+2=4 \equiv 1$.
Choices for $(a,b,c)$:
– Two from $R_1$, One from $R_2$.
– Ways: $3 \text{ (positions for } R_2) \times 1 \times 1 \times 2 \text{ (choice from } R_2) = 6$.
3. $(0, 2, 2) \to 0+2+2=4 \equiv 1$.
Choices for $(a,b,c)$:
– One from $R_0$, Two from $R_2$.
– Ways: $3 \text{ (positions for } R_0) \times 2 \text{ (choice from } R_0) \times 2 \times 2 \text{ (choices from } R_2) = 24$.
4. $(2, 2, ?)$ No. $2+2+2=6 \equiv 0$.
5. $(1, 1, 1) \to 3 \equiv 0$. No.

Total ways = $12 + 6 + 24 = 42$.
Check constraint “greater than 5000”: The smallest number formed is 5000. But digits must be from $\{0,1,2,5,9\}$. 5000 uses 5,0,0,0. Sum=5 (Not div by 3). So 5000 is not counted.
Check constraint “less than 9000”: First digit is fixed at 5, so all are $<6000 < 9000$. So all formed numbers are valid.

Ans. (42)