Maths Q6 – JEE-MAIN 2026

Home Maths Permutations and CombinationsQuestion #313

The number of seven-digit numbers, that can be formed by using the digits

1, 2, 3, 5

and

7

such that each digit is used at least once, is :

Clue 1 of 3: Identify the structure of the problem

You need to form a 7-digit number using 5 distinct digits, with each digit used at least once. This implies some digits must be repeated.

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Step 1: Determine the repetition pattern of digits✦ Active

We need to form a 7-digit number using 5 distinct digits (1, 2, 3, 5, 7) such that each digit is used at least once. This means 5 digits are used once, and the remaining $7 - 5 = 2$ positions must be filled by repeating some of these 5 digits. There are two possible scenarios for these repetitions:

Case 1: One digit is repeated three times, and the other four digits are used once. (e.g., $d_{1}, d_{1}, d_{1}, d_{2}, d_{3}, d_{4}, d_{5}$ )

Case 2: Two distinct digits are repeated twice each, and the other three digits are used once. (e.g., $d_{1}, d_{1}, d_{2}, d_{2}, d_{3}, d_{4}, d_{5}$ )

Step 2: Calculate arrangements for Case 1○ Expand

In this case, one digit is repeated three times. The number of ways to choose which of the 5 digits is repeated three times is $(\binom{5}{1}) = 5$ .

For each choice (e.g., if '1' is repeated), the set of digits is {1, 1, 1, 2, 3, 5, 7}. The number of distinct permutations of these 7 digits is given by the multinomial coefficient formula:

\frac{7!}{3! 1! 1! 1! 1!} = \frac{7!}{3!} = \frac{5040}{6} = 840

Total for Case 1: $5 \times 840 = 4200$ .

Step 3: Calculate arrangements for Case 2 and find the total○ Expand

In this case, two distinct digits are repeated twice each. The number of ways to choose which two of the 5 digits are repeated twice is $(\binom{5}{2}) = \frac{5 \times 4}{2} = 10$ .

For each choice (e.g., if '1' and '2' are repeated), the set of digits is {1, 1, 2, 2, 3, 5, 7}. The number of distinct permutations of these 7 digits is:

\frac{7!}{2! 2! 1! 1! 1!} = \frac{7!}{2! 2!} = \frac{5040}{4} = 1260

Total for Case 2: $10 \times 1260 = 12600$ .

The total number of seven-digit numbers is the sum of arrangements from both cases: $4200 + 12600 = 16800$ .

💡 Teacher's Secret Hint

Ensure you consider all possible ways to distribute the 'extra' repetitions among the available digits.

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Maths Question 6 – JEE-MAIN 2026

Clue 1 of 3: Identify the structure of the problem

🧠 Full Solution Path