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Exercise 6.2 · Q13

Q.How many numbers are there between 100 and 1000 such that 7 is in the units place.

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Among the 3-digit numbers between 100 and 1000, fixing the units digit at 7 and letting the hundreds and tens digits vary freely gives 9×10=909\times10=90 numbers.

[!FORMULA] By the multiplication principle, the count of 3-digit numbers with a fixed units digit is (choices for hundreds digit) ×\times (choices for tens digit) ×1\times 1 (units digit fixed), where the hundreds digit ranges over {1,…,9}\{1,\dots,9\} (cannot be 0) and the tens digit ranges over {0,…,9}\{0,\dots,9\}.

  1. Numbers between 100 and 1000 are exactly the 3-digit numbers (100 to 999).

  2. The units digit is fixed at 7 (only 1 choice).

  3. The tens digit can be any digit from 0 to 9: 1010 choices.

  4. The hundreds digit can be any digit from 1 to 9 (it cannot be 0, since that would make it a 2-digit number): 99 choices.

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