how many 3 digit positive numbers have their hundreds, tens, and units digits in strictly ascending order

Let the digit in hundreds place be n, since its is strictly ascending order, the digit in the tens place is n + 1 and the digit in the units place is n + 2.

So, the digits are n, n + 1, n + 2.

The positive 3 digit numbers

So, starting with n = 1, we find the number of 3 digit positive integers in ascending order.

So, when n = 1,

n, n + 1, n + 2.

1, 1 + 1, 1 + 2.

1, 2, 3

So, the first number is 123

So, when n = 2,

n, n + 1, n + 2.

2, 2 + 1, 2 + 2.

2, 3, 4

So, the second number is 234

So, when n = 3,

n, n + 1, n + 2.

3, 3 + 1, 3 + 2.

3, 4, 5

So, the third number is 345

So, when n = 4,

n, n + 1, n + 2.

4, 4 + 1, 4 + 2.

4, 5, 6

So, the fourth number is 456

So, when n = 5,

n, n + 1, n + 2.

5, 5 + 1, 5 + 2.

5, 6, 7

So, the fifth number is 567

So, when n = 6,

n, n + 1, n + 2.

6, 6 + 1, 6 + 2.

6, 7, 8

So, the sixth number is 678

When n = 7,

n, n + 1, n + 2.

7, 7 + 1, 7 + 2.

7, 8, 9

So, the seventh number is 789

When n = 8,

n, n + 1, n + 2.

8, 8 + 1, 8 + 2.

8, 9, 10

Since we cannot have 10 in the last digit, we ignore this number.

The 3 digit positive numbers

So, the numbers are 123, 234, 345, 456, 567, 678, and 789

So, there are seven 3 digit positive numbers that have their hundreds, tens, and units digits in strictly ascending order

Learn more about positive numbers here:

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