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(C) To find the number of positive integers less than $10000$ that contain the digit $5$ at least once,we use the complement method.Total positive int

Vedclass · Accepted Answer

$3439$

Vedclass · Answer

(C) To find the number of positive integers less than $10000$ that contain the digit $5$ at least once,we use the complement method.Total positive integers less than $10000$ are $9999$.We calculate the number of integers that do not contain the digit $5$ at all.These integers can be represented as $d_1 d_2 d_3 d_4$ where each digit $d_i \in \{0, 1, 2, 3, 4, 6, 7, 8, 9\}$.There are $9$ choices for each position (excluding $5$).For a $4$-digit representation (including leading zeros for numbers less than $1000$),there are $9 	imes 9 	imes 9 	imes 9 = 9^4 = 6561$ such combinations.Since we are looking for positive integers,we exclude the case where all digits are $0$ $(0000)$,so there are $6561 - 1 = 6560$ integers that do not contain the digit $5$.The number of integers containing at least one $5$ is $9999 - 6560 = 3439$.

The number of positive integers less than $10000$ which contain the digit $5$ at least once is

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