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 :
- 15400
- 17800
- 16800
- 29400
Answer
(3) 16800
Solution
All 5 digits must appear in 7 places, so 2 places are extra, which gives two cases.
Case I: one digit is used 3 times and the other four once each. The digit can be chosen in 5 ways and the arrangements are 7!3! = 840, so
5 840 = 4200
Case II: two digits are used twice each and the other three once each. The two digits can be chosen in 5C2 = 10 ways and the arrangements are 7!2! 2! = 1260, so
10 1260 = 12600
∴ Total = 4200 + 12600 = 16800
Answer: (3)