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Exercise 6.2 · Q7

Q.Find the number of three digit even positive integers.

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Counting 3-digit even numbers by filling hundreds, tens, and units digits with the multiplication principle gives 9×10×5=4509\times10\times5=450.

[!FORMULA] By the multiplication principle, the number of 3-digit numbers is the product of the number of valid choices for each place value: hundreds digit ∈{1,…,9}\in\{1,\dots,9\} (cannot be 00, else it would not be a 3-digit number), tens digit ∈{0,…,9}\in\{0,\dots,9\}, and — for an even number — units digit ∈{0,2,4,6,8}\in\{0,2,4,6,8\}.

  1. Hundreds place: any digit from 1 to 9 (cannot be 0), giving 99 choices.

  2. Tens place: any digit from 0 to 9, giving 1010 choices (no restriction, and repetition of digits across places is allowed since nothing forbids it).

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