This is a counting problem, not a construction problem, so we can approach it using recursion. Since the problem has two natural parts, looking from the left and looking from the right, break it up and solve for just one part first.

Let `b(N, L, R)`

be the number of solutions, and let `f(N, L)`

be the number of arrangements of `N`

blocks so that `L`

are visible from the left. First think about `f`

because it’s easier.

*APPROACH 1*

Let’s get the initial conditions and then go for recursion. If all are to be visible, then they must be ordered increasingly, so

```
f(N, N) = 1
```

If there are suppose to be more visible blocks than available blocks, then nothing we can do, so

```
f(N, M) = 0 if N < M
```

If only one block should be visible, then put the largest first and then the others can follow in any order, so

```
f(N,1) = (N-1)!
```

Finally, for the recursion, think about the position of the tallest block, say `N`

is in the `k`

th spot from the left. Then choose the blocks to come before it in `(N-1 choose k-1)`

ways, arrange those blocks so that exactly `L-1`

are visible from the left, and order the `N-k`

blocks behind `N`

it in any you like, giving:

```
f(N, L) = sum_{1<=k<=N} (N-1 choose k-1) * f(k-1, L-1) * (N-k)!
```

In fact, since `f(x-1,L-1) = 0`

for `x<L`

, we may as well start `k`

at `L`

instead of `1`

:

```
f(N, L) = sum_{L<=k<=N} (N-1 choose k-1) * f(k-1, L-1) * (N-k)!
```

Right, so now that the easier bit is understood, let’s use `f`

to solve for the harder bit `b`

. Again, use recursion based on the position of the tallest block, again say `N`

is in position `k`

from the left. As before, choose the blocks before it in `N-1 choose k-1`

ways, but now think about each side of that block separately. For the `k-1`

blocks left of `N`

, make sure that exactly `L-1`

of them are visible. For the `N-k`

blocks right of `N`

, make sure that `R-1`

are visible and then reverse the order you would get from `f`

. Therefore the answer is:

```
b(N,L,R) = sum_{1<=k<=N} (N-1 choose k-1) * f(k-1, L-1) * f(N-k, R-1)
```

where `f`

is completely worked out above. Again, many terms will be zero, so we only want to take `k`

such that `k-1 >= L-1`

and `N-k >= R-1`

to get

```
b(N,L,R) = sum_{L <= k <= N-R+1} (N-1 choose k-1) * f(k-1, L-1) * f(N-k, R-1)
```

*APPROACH 2*

I thought about this problem again and found a somewhat nicer approach that avoids the summation.

If you work the problem the opposite way, that is think of adding the smallest block instead of the largest block, then the recurrence for `f`

becomes much simpler. In this case, with the same initial conditions, the recurrence is

```
f(N,L) = f(N-1,L-1) + (N-1) * f(N-1,L)
```

where the first term, `f(N-1,L-1)`

, comes from placing the smallest block in the leftmost position, thereby adding one more visible block (hence `L`

decreases to `L-1`

), and the second term, `(N-1) * f(N-1,L)`

, accounts for putting the smallest block in any of the `N-1`

non-front positions, in which case it is not visible (hence `L`

stays fixed).

This recursion has the advantage of always decreasing `N`

, though it makes it more difficult to see some formulas, for example `f(N,N-1) = (N choose 2)`

. This formula is fairly easy to show from the previous formula, though I’m not certain how to derive it nicely from this simpler recurrence.

Now, to get back to the original problem and solve for `b`

, we can also take a different approach. Instead of the summation before, think of the visible blocks as coming in packets, so that if a block is visible from the left, then its packet consists of all blocks right of it and in front of the next block visible from the left, and similarly if a block is visible from the right then its packet contains all blocks left of it until the next block visible from the right. Do this for all but the tallest block. This makes for `L+R`

packets. Given the packets, you can move one from the left side to the right side simply by reversing the order of the blocks. Therefore the general case `b(N,L,R)`

actually reduces to solving the case `b(N,L,1) = f(N,L)`

and then choosing which of the packets to put on the left and which on the right. Therefore we have

```
b(N,L,R) = (L+R choose L) * f(N,L+R)
```

Again, this reformulation has some advantages over the previous version. Putting these latter two formulas together, it’s much easier to see the complexity of the overall problem. However, I still prefer the first approach for constructing solutions, though perhaps others will disagree. All in all it just goes to show there’s more than one good way to approach the problem.

*What’s with the Stirling numbers?*

As Jason points out, the `f(N,L)`

numbers are precisely the (unsigned) Stirling numbers of the first kind. One can see this immediately from the recursive formulas for each. However, it’s always nice to be able to see it directly, so here goes.

The (unsigned) Stirling numbers of the First Kind, denoted `S(N,L)`

count the number of permutations of `N`

into `L`

cycles. Given a permutation written in cycle notation, we write the permutation in canonical form by beginning the cycle with the largest number in that cycle and then ordering the cycles increasingly by the first number of the cycle. For example, the permutation

```
(2 6) (5 1 4) (3 7)
```

would be written in canonical form as

```
(5 1 4) (6 2) (7 3)
```

Now drop the parentheses and notice that if these are the heights of the blocks, then the number of visible blocks from the left is exactly the number of cycles! This is because the first number of each cycle blocks all other numbers in the cycle, and the first number of each successive cycle is visible behind the previous cycle. Hence this problem is really just a sneaky way to ask you to find a formula for Stirling numbers.