How many four-digit even numbers are possible if the leftmost digit cannot be zero

In this problem, we apply the Fundamental Counting Principle. This is how it is used. If there are 'm' things of object A and there are 'n' things of object B, then the total number of ways of arranging A and B is m×n. 

Now, we have a four-digit number. So, we are going to divide four numbers. We know that there are 10 1-digit numbers in a 4-digit number. In the 10 digits, one is zero, and the other 9 are non-zeros. For the rest of the 3 digits, there are no restrictions. It is said that the leftmost side cannot be zero, so that is 9. So, assuming that the numbers can be repeated, the answer would be 9×10×10×10 = 9,000 ways. There are a total of 9,000 ways.

Now, if we assume that the numbers don't repeat. The answer would be 9×9×8×7 = 4,536 ways. You will have to subtract 1 number to the next multiplied number to eliminate repeating the same numbers.

B. because the 7 represents "seven times", x represents "a number", 12 represents "plus twelve", and -4 represents "minus four"