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Bunuel
How many three-digit positive integers with different digits are greater than 700?

A. 180
B. 216
C. 243
D. 300
E. 310


Consider all positive integers with three different digits. (Note that zero cannot be the first digit.) Find the number of them which are: (a) greater than 700; (b) odd; (c) divisible by 5

You have the 10s and units digits to play with in this problem. For numbers between 700 to 799, the tens digit can be anything but 7 and the units digit can be anything but 7 and the number forming the 10s digit. So, you had 9 possible numbers in the 10s digit (0 to 9 except 7) and 8 possible numbers in the units digit (0 to 9 except 7 and the number in the 10s digit). Hence, you have 9 x 8 = 72 possible numbers.

Similarly, for 800 to 899 and from 900 to 999, you have 72 possible numbers each.

In total , 72 + 72 + 72 = 216 numbers

Answer: option B
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