The number of seven-digit numbers that can be | vedclass

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(C) We have $5$ distinct digits to fill $7$ positions, and each digit must appear at least once.This implies that two digits must be repeated.The poss

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$16800$

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(C) We have $5$ distinct digits to fill $7$ positions, and each digit must appear at least once.This implies that two digits must be repeated.The possible partitions of $7$ into $5$ parts are $(3, 1, 1, 1, 1)$ and $(2, 2, 1, 1, 1)$.Case $1$: Partition $(3, 1, 1, 1, 1)$First, choose the digit that appears $3$ times: $\binom{5}{1} = 5$ ways.Then, arrange these $7$ digits: $\frac{7!}{3!1!1!1!1!} = \frac{5040}{6} = 840$ ways.Total for Case $1 = 5 	imes 840 = 4200$.Case $2$: Partition $(2, 2, 1, 1, 1)$First, choose the $2$ digits that appear twice each: $\binom{5}{2} = 10$ ways.Then, arrange these $7$ digits: $\frac{7!}{2!2!1!1!1!} = \frac{5040}{4} = 1260$ ways.Total for Case $2 = 10 	imes 1260 = 12600$.Grand Total = $4200 + 12600 = 16800$.

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:

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