How many 7 digit numbers can be formed with the first two digits odd and the last five digits even if repition of digits is allowed?

Answer 1

There are 5 choices for every digit.

Each of the first two can be any one of (1, 3, 5, 7, or 9).

Each of the last five can be any one of (2, 4, 6, 8, or 0).

The total number of possibilities is (5 x 5 x 5 x 5 x 5 x 5 x 5) = 57 = 78,125

This is only 0.78125% of the 10 million possible numbers if all 10 digits were allowed in any position.