Find the number of all 10-digit integers in which no two consecutive digits are the same.?

The first digit can be any of the nine non-zero digits.
The second digit can be any digit EXCEPT for whatever the first one is. So that leaves 10-1 = 9 possibilities.
The third digit can be any digit except for whatever the second digit is. So that has 9 possibilities.
The same logic goes for digits in the 4th through 10th place. So that gives us:

9*9*9*9*9 * 9*9*9*9*9 = 9^10 = 3,486,784,401