Eight policemen are to be posted to guard three separate banks. If no bank is to be guarded by less than two policemen, in how many ways can they be posted?
We can solve this question by using stars and bars.
First, let's consider the scenario where we don't have any restrictions on the number of policemen guarding each bank. We can use stars and bars to distribute the policemen among the three banks. Then, each bank is represented by a "bar," and we have 8 "stars" to distribute among the three banks.
We have 8 stars and 2 bars. We can place the 2 bars anywhere among the 8+2=10 slots (8 stars and 2 bars) in $\binom{10}{2} = 45$ ways.
However, in some of these distributions, a bank may end up with fewer than 2 policemen. We need to calculate the number of distributions where at least one bank is guarded by less than 2 policemen.
First, let's find the number of distributions in which only one bank is guarded by less than 2 policemen. There are three possible banks that can be guarded by fewer policemen. We can assign 7 policemen in ${8 \choose 7} = 8$ ways, and the remaining bank will be assigned the remaining 1 policeman.
Thus, there are $3 \times 8 = 24$ distributions where only one bank is guarded by less than 2 policemen.
Next, let's find the number of distributions in which two banks are guarded by less than 2 policemen. There are three possible combinations of two banks that can be guarded by fewer policemen. We can assign 6 policemen in ${8 \choose 6} = 28$ ways to these two banks, and the remaining bank will be assigned the remaining 2 policemen.
Thus, there are $3 \times 28 = 84$ distributions where two banks are guarded by less than 2 policemen.
Finally, let's find the number of distributions in which all three banks are guarded by less than 2 policemen. In this case, there can only be one policeman assigned to each bank. We can assign the policemen in $1$ way.
Therefore, there are $1 + 24 + 84 = 109$ distributions where at least one bank is guarded by less than 2 policemen.
However, we need to subtract these distributions from the total number of distributions to get the final answer. Therefore, there are $45 - 109 = \boxed{-64}$ ways.