**how many non-negative integers less than 104 that contain the digit 2 ?**

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**Here's the Solution to this Question**

We identify one-digit number $a$ with $000a,$ two-digit number $n=\overline{ab}$ with $00ab,$ and three-digit number $n=\overline{abc}$ with $0abc.$ Taking into account that there are ten digits, we conclude by Combinatorial product rule that the number of non-negative integers less than $10^4$ is equal to $10\cdot 10\cdot 10\cdot 10=10,000.$ Since without digit 2 there are nine digits, we conclude by Combinatorial product rule that the number of non-negative integers less than $10^4$ that do not contain the digit 2 is equal to $9\cdot 9\cdot 9\cdot 9=6,561.$

We conclude that the number of non-negative integers less than 104 that contain the digit 2 is equal to $10,000-6,561=3,439.$