Answer:
It gives you 28 different 10-digit numbers
Step-by-step explanation:
The leftmost position digit can not be 0 --- hence, it is 1, since there is no other possibilities.
The rightmost position (the "ones" position) must be 0 to provide an even number.
Hence, you have 8 remaining positions to fill with six ones and two zeros by an arbitrary way.
You can place two digits of zero among 8 positions by =
different ways = 4*7 = 28 different ways,