or b) **6*

Note that a “four-digit” number does not begin with a “0” [that would make it a three-digit number].

For Case (a), there are 5 groups that look like: 16**, 26**, 36**, 46**, and 56**. Each of these may have the last two digits be 54, 53, 52, 51, 50, 43, 42, 41, 40, 32, 31, 30, 21, 20, or 10. That makes 5*15 = 75 possible four-digit up-and-down numbers with a maximum digit of 6.

What about a maximum digit of 5? 15**, 25**, 35**, and 45**. Each of these can have the last two digits be 43, 42, 41, 40, 32, 31, 30, 21, 20, or 10. That is 4*10=40 more.

What about a maximum digit of 4? 14**, 14**, and 34**. Each of these can have the last two digits be 32, 31, 30, 21, 20, or 10. So, 3*6=18 more.

And, a maximum digit of 3? 13** and 23**. Each of these can have the last two digits be 21, 20, or 10. That’s 6 more.

Can the maximum digit be 2? Yes: 1210. That’s 1 last value.

75 + 40 + 18 + 6 + 1 = 140

Now, Case (b):

The groups with 5 as a maximum digit look like: **54, **53, **52, **51 and **50. Each of these groups may have digits of 12, 13, 14, 23, 24, and 34. How many more possibilities does that create?

What about groups with 4 as a maximum digit? **43, **42, **41, and **40. Each of these groups may have 12, 13, and 23 as the first two digits. How many more is that?

Finally, 3 may be the maximum digit. The groups look like **32, **31, and **30. Each of these may have digits of 12. How many more is that?

Now, add up all the values. You should get

**245**.