Number of seven-digit number formed by using the digits 1, 2, 3, 4, 5 at least once are

The correct option is 16800.

To find the number of seven-digit numbers formed using the digits 1, 2, 3, 4, and 5 such that each digit is used at least once, we can break down the problem into different cases based on how the digits are repeated.

Since we are forming a 7-digit number using 5 distinct digits, and each digit must appear at least once, we have two possible cases for the frequency distribution of the digits:

Case 1: One digit is repeated 3 times, and the remaining four digits appear 1 time each.
The frequency distribution of the digits is {3, 1, 1, 1, 1}.
1. Choose which of the 5 digits will be repeated 3 times:
$C_1^5=5$ ways.
2. Arrange these 7 digits (where one digit is repeated 3 times and the other four are unique):
$\frac{7!}{3!}=\frac{5040}{6}=840$ ways.
3. Total numbers formed in Case 1:
$5\times 840=4200$.

Case 2: Two digits are repeated 2 times each, and the remaining three digits appear 1 time each.
The frequency distribution of the digits is {2, 2, 1, 1, 1}.
1. Choose which 2 digits out of the 5 will be repeated 2 times:
$C_2^5=10$ ways.
2. Arrange these 7 digits (where two digits are repeated 2 times each and the other three are unique):
$\frac{7!}{2!2!}=\frac{5040}{4}=1260$ ways.
3. Total numbers formed in Case 2:
$10\times 1260=12600$.

Total Number of Ways:
Summing the results from both cases, we get:
$\text{Total ways}=4200+12600=16800$.