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

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Goober

Mathematics

27 March, 19:31

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

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Gianna Camacho · Answer 1

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.