Math JEE Adv. 2025 paper 1

Math JEE Adv. 2025 paper 1

on July 09, 2025

To solve the problem, we'll count the number of seven-digit numbers formed using digits {0, 1, 2} where at least one of the digits 0 or 1 appears exactly twice. Here's the step-by-step solution:

Step 1: Total Possible Numbers

Each of the 7 digits can be 0, 1, or 2.
The first digit cannot be 0 (to ensure it's a seven-digit number).
Total numbers: $2 \times 3^{6} = 2 \times 729 = 1458$ .

Step 2: Numbers Where 0 Appears Exactly Twice

Choose positions for 0: $(\binom{6}{2})$ (since the first digit cannot be 0).
Fill the remaining 5 digits: Each can be 1 or 2 (since we're fixing exactly two 0s).
Count: $(\binom{6}{2}) \times 2^{5} = 15 \times 32 = 480$ .

Step 3: Numbers Where 1 Appears Exactly Twice

Choose positions for 1: $(\binom{7}{2})$ .
Fill the remaining 5 digits: Each can be 0 or 2 (but the first digit cannot be 0).
- If the first digit is 1: $(\binom{6}{1}) \times 2^{5} = 6 \times 32 = 192$ .
- If the first digit is not 1 (i.e., it's 2): $(\binom{6}{2}) \times 2^{4} = 15 \times 16 = 240$ .
Total for 1 appearing exactly twice: $192 + 240 = 432$ .

Step 4: Numbers Where Both 0 and 1 Appear Exactly Twice

Overlap count: Numbers where both 0 and 1 appear exactly twice.
Choose positions for 0: $(\binom{6}{2})$ .
Choose positions for 1 from remaining: $(\binom{5}{2})$ .
Fill the remaining 3 digits: Must be 2.
Count: $(\binom{6}{2}) \times (\binom{5}{2}) = 15 \times 10 = 150$ .

Step 5: Apply Inclusion-Exclusion Principle

Total desired numbers: Numbers where 0 appears exactly twice + Numbers where 1 appears exactly twice - Numbers where both appear exactly twice.
Calculation: $480 + 432 - 150 = 762$ .

Final Answer

The number of such seven-digit numbers is $762$ .

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