Problem 154

To solve this problem, we can consider the number of selection methods that include key project A, the number of selection methods that include general project B, and then subtract the selection methods that include both A and B.

First, let's calculate the selection methods that include key project A:

1. Choose project A from the key projects. There is $1$ way to do this.

2. Choose $1$ more key project from the remaining $3$ key projects. There are $\binom{3}{1} = 3$ ways to do this.

3. Choose $2$ general projects from $6$. There are $\binom{6}{2} = 15$ ways to do this.

Therefore, the number of ways to include project A is $1 \times 3 \times 15 = 45$.

Next, we calculate the selection methods that include general project B:

1. Choose project B from the general projects. There is $1$ way to do this.

2. Choose $1$ more general project from the remaining $5$ general projects. There are $\binom{5}{1} = 5$ ways to do this.

3. Choose $2$ key projects from $4$. There are $\binom{4}{2} = 6$ ways to do this.

Therefore, the number of ways to include project B in the general category is $1 \times 5 \times 6 = 30$.

Now, let's subtract the ways in which both A and B are selected:

1. Choose project A from the key projects. There is $1$ way to do this.

2. Choose project B from the general projects. There is $1$ way to do this.

3. Choose $1$ key project from the remaining $3$ key projects. There are $\binom{3}{1} = 3$ ways to do this.

4. Choose $1$ general project from the remaining $5$ general projects. There are $\binom{5}{1} = 5$ ways to do this.

Therefore, the number of ways to include both project A and project B is $1 \times 1 \times 3 \times 5 = 15$.

To find the total number of ways, we combine the selections that include A or B but subtract the overlapping cases where both are included: $45 + 30 - 15$.

Calculation:
$$45 + 30 - 15 = 60$$

Hence, the number of different selection methods such that either key project A or general project B or both are selected is $\boxed{60}$.