How many three digits numbers N=abc are there such that a≤b and c≤b?

To find the number of three-digit numbers N=abc such that a≤b and c≤b, we need to consider the possible values for a, b, and c.

1. For a: Since a≤b, the possible values for a are 1, 2, 3, ..., 9.

2. For b: Since b is greater than or equal to a, the possible values for b depend on the value of a. If a=1, then b can be 1, 2, ..., 9. If a=2, then b can be 2, ..., 9. And so on, until a=9, where b can only be 9.

3. For c: Since c≤b, the possible values for c also depend on the value of b. For each value of b, the possible values for c can be 0, 1, 2, ..., b.

To find the total number of three-digit numbers satisfying these conditions, we need to count the possibilities for each variable and sum them up.

For example, let's consider the case where a=1:
- a=1, b can be 1, 2, ..., 9.
- For each value of b, c can be 0, 1, 2, ..., b.

So, for a=1, there are 9 possible values for b and the number of possibilities for c depends on the value of b.

We do the same analysis for each possible value of a (from 1 to 9) and add up the results.

To get the final answer, we sum up the number of possibilities for each case.

Let's calculate it step by step:
For a=1, there are 9 possibilities for b.
- For b=1, there is 1 possibility for c (0).
- For b=2, there are 2 possibilities for c (0, 1).
- For b=3, there are 3 possibilities for c (0, 1, 2).
- And so on, up to b=9, where there are 9 possibilities for c (0, 1, 2, ..., 8, 9).

So, for a=1, there are 9 + 1 + 2 + 3 + ... + 9 possibilities for three-digit numbers.

We can calculate this sum by using the arithmetic series formula: Sn = n/2 * (a + l), where n is the number of terms, a is the first term, and l is the last term.

Using the formula, we have:
S9 = 9/2 * (1 + 9) = 9/2 * 10 = 45

So, for a=1, there are 45 possibilities for three-digit numbers.

We repeat the same calculations for each possible value of a (from 1 to 9) and add up the results:

For a=1, there are 45 possibilities.
For a=2, there are 36 possibilities.
For a=3, there are 28 possibilities.
For a=4, there are 21 possibilities.
For a=5, there are 15 possibilities.
For a=6, there are 10 possibilities.
For a=7, there are 6 possibilities.
For a=8, there are 3 possibilities.
For a=9, there is 1 possibility.

Adding up these possibilities, we get:
45 + 36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 = 165

So, there are 165 three-digit numbers N=abc such that a≤b and c≤b.