Question #293424

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


Expert's answer

We identify one-digit number aa with 000a,000a, two-digit number n=ab‾n=\overline{ab} with 00ab,00ab, and three-digit number n=abc‾n=\overline{abc} with 0abc.0abc. Taking into account that there are ten digits, we conclude by Combinatorial product rule that the number of non-negative integers less than 10410^4 is equal to 10⋅10⋅10⋅10=10,000.10\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 10410^4 that do not contain the digit 2 is equal to 9⋅9⋅9⋅9=6,561.9\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=3,439.10,000-6,561=3,439.


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