Permutations & CombinationsmediumPYQ · JEE Main · 2 Apr 2026 · Shift 1 (Morning)Free

Seven-Digit Numbers Using 5 Digits Each at Least Once: 16800 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
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
A1540015400
B1780017800
C1680016800correct
D2940029400
Solution
Step 1: Seven positions must be filled by the five digits with each used at least once, so exactly two "extra" repeats are distributed. Two patterns are possible. Step 2: One digit used three times (pattern 3,1,1,1,1): choose which digit repeats in 5C1{}^5C_1 ways, then arrange:
5C1×7!3!=5×840=4200.{}^5C_1\times\frac{7!}{3!}=5\times840=4200.
 Step 3: Two digits used twice each (pattern 2,2,1,1,1): choose the two repeating digits in 5C2{}^5C_2 ways, then arrange:
5C2×7!2! 2!=10×1260=12600.{}^5C_2\times\frac{7!}{2!\,2!}=10\times1260=12600.
 Step 4: Total number of seven-digit numbers:
4200+12600=16800.4200+12600=16800.
 Correct answer: (3)
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