How many positive three-digit integers have the property that the tens digit is greater than or equal to the hundreds digit and the ones digit is greater than or equal to the tens digit?

Please help :)
thanks
-Zach

9a

Let a = hundreds, b = tens, c = ones

a ≤ b ≤ c

Examples of what you are seeking are:

111
112
113...

122
123
124...

133
134
135...

222
223
224...

...999

Unfortunately, I cannot think of an equation to help you solve this problem. Sorry.

To find the number of three-digit integers that satisfy the given conditions, we can break down the problem into three steps:

Step 1: Determine the possible values for the hundreds digit.
Since the hundreds digit must be less than or equal to the tens digit, there are 10 possible values for the hundreds digit: 1, 2, 3, ..., 9, 0. (0 is also included because we are looking for positive integers.)

Step 2: Determine the possible values for the tens digit.
Since the tens digit must be greater than or equal to the hundreds digit, there are 10 possible values for the tens digit as well. If the hundreds digit is 1, the tens digit can be 1, 2, 3, ..., 9, 0. If the hundreds digit is 2, the tens digit can be 2, 3, ..., 9, 0. And so on.

Step 3: Determine the possible values for the ones digit.
Similarly, the ones digit must be greater than or equal to the tens digit, so there are 10 possible values for the ones digit.

Now, to find the number of three-digit integers that satisfy the given conditions, we multiply the number of possible values for each digit together: 10 × 10 × 10 = 1000.

However, we need to exclude the case where the three-digit integer is equal to 1000, which is not a positive three-digit integer. Therefore, the final answer is 1000 - 1 = 999.

So, there are 999 positive three-digit integers that have the property described.