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JEE Main 20262 AprilMorning Shift
Exam
JEE Main
Year
2026
Date
2 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Permutation and Combination

JEE Main 2026 · 2 April · Morning Shift · Mathematics · Permutation and Combination

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

The number of seven-digit numbers, that can be formed by using the digits 1,2,3,51, 2, 3, 5 and 77 such that each digit is used at least once, is :

Options

  1. A1540015400
  2. B1780017800
  3. C1680016800
  4. D2940029400

Correct answer

C. 1680016800

Step-by-step solution

A seven-digit number is to be formed using the 55 digits 1,2,3,5,1, 2, 3, 5, and 77. Since each digit must be used at least once, the remaining 22 places can be filled by either repeating one digit two more times, or repeating two digits one more time each.

This gives rise to two cases for the frequencies of the digits:

Case 1: One digit appears 33 times and the other four digits appear 11 time each.
The number of ways to choose the digit that appears 33 times is 5C1^{5}C_{1}.
The number of arrangements of these 77 digits is 7!3!\dfrac{7!}{3!}.
Number of numbers formed in this case = 5C1×7!3!=5×50406=4200^{5}C_{1} \times \dfrac{7!}{3!} = 5 \times \dfrac{5040}{6} = 4200.

Case 2: Two digits appear 22 times each and the other three digits appear 11 time each.
The number of ways to choose the two digits that appear 22 times is 5C2^{5}C_{2}.
The number of arrangements of these 77 digits is 7!2!2!\dfrac{7!}{2!2!}.
Number of numbers formed in this case = 5C2×7!2!2!=10×50404=12600^{5}C_{2} \times \dfrac{7!}{2!2!} = 10 \times \dfrac{5040}{4} = 12600.

Total number of seven-digit numbers = 4200+12600=168004200 + 12600 = 16800.

Answer: 1680016800
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