Seven Digit Numbers Formed Using Specific Digits at Least Once

To find the total number of seven-digit numbers that can be formed using the digits $\{1, 2, 3, 5, 7\}$ such that each digit is used at least once, we can use the principles of combinations and permutations.

Since there are $5$ distinct available digits and we need to construct a $7$-digit number where each digit appears at least once, the distribution of the frequencies of the $5$ digits across the $7$ positions can occur in two distinct cases:

Case 1: One digit is repeated three times, and the remaining four digits appear once each.

• Frequency distribution: $(3, 1, 1, 1, 1)$
• Select the digit that repeats 3 times: $\binom{5}{1} = 5$ ways.
• Arrange the 7 digits: The number of ways to arrange $7$ digits where one digit repeats $3$ times is: $\frac{7!}{3!} = \frac{5040}{6} = 840$
• Total for Case 1: $N_1 = 5 \times 840 = 4200$

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Case 2: Two digits are repeated two times each, and the remaining three digits appear once each.

• Frequency distribution: $(2, 2, 1, 1, 1)$
• Select the 2 digits that repeat 2 times each: $\binom{5}{2} = 10$ ways.
• Arrange the 7 digits: The number of ways to arrange $7$ digits where two digits repeat $2$ times each is: $\frac{7!}{2! \times 2!} = \frac{5040}{4} = 1260$
• Total for Case 2: $N_2 = 10 \times 1260 = 12600$

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Alternative Method (Principle of Inclusion-Exclusion):

The total number of onto functions from a set of $7$ positions to $5$ digits is given by: $\text{Total} = \sum_{j=0}^{5} (-1)^{5-j} \binom{5}{j} j^7$ $\text{Total} = 5^7 - \binom{5}{1} 4^7 + \binom{5}{2} 3^7 - \binom{5}{3} 2^7 + \binom{5}{4} 1^7$ $\text{Total} = 78125 - 5(16384) + 10(2187) - 10(128) + 5(1)$ $\text{Total} = 78125 - 81920 + 21870 - 1280 + 5 = 16800$

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Total Number of Seven-Digit Numbers:

$\text{Total} = N_1 + N_2 = 4200 + 12600 = 16800$

Hence, the correct option is C.