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(C) The given digits are $0, 1, 2, 3$. We need to form distinct positive integers using these digits at most once.Case $I$: $4$-digit integers.The fir

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$48$

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(C) The given digits are $0, 1, 2, 3$. We need to form distinct positive integers using these digits at most once.Case $I$: $4$-digit integers.The first place can be filled by any of the $3$ non-zero digits $(1, 2, 3)$. The remaining $3$ places can be filled by the remaining $3$ digits in $3 	imes 2 	imes 1$ ways.Number of $4$-digit integers $= 3 	imes 3 	imes 2 	imes 1 = 18$.Case $II$: $3$-digit integers.The first place can be filled by $3$ choices $(1, 2, 3)$. The remaining $2$ places can be filled by the remaining $3$ digits in $3 	imes 2$ ways.Number of $3$-digit integers $= 3 	imes 3 	imes 2 = 18$.Case $III$: $2$-digit integers.The first place can be filled by $3$ choices $(1, 2, 3)$. The second place can be filled by the remaining $3$ digits in $3$ ways.Number of $2$-digit integers $= 3 	imes 3 = 9$.Case $IV$: $1$-digit integers.The possible integers are $1, 2, 3$.Number of $1$-digit integers $= 3$.Total number of distinct positive integers $= 18 + 18 + 9 + 3 = 48$.

The number of distinct positive integers that can be formed using $0, 1, 2, 3$,where each integer is used at most once,is equal to

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