Problem:
In how many ways can a selection of 6 out of 4 teachers and 8 students be done so as to include at least 2 teachers?

Solution:
To solve this problem, we can consider two cases: one where we select exactly 2 teachers and another where we select more than 2 teachers.

Case 1: Selecting exactly 2 teachers
In this case, we need to select 2 teachers and 4 students from the given pool. The number of ways to select 2 teachers out of 4 is given by the combination formula:

C(4, 2) = 4! / (2! * (4 - 2)!) = 6

Similarly, the number of ways to select 4 students out of 8 is:

C(8, 4) = 8! / (4! * (8 - 4)!) = 70

Since these two selections are independent, we can multiply the number of ways to select teachers and students:

6 * 70 = 420

Therefore, there are 420 ways to select 6 individuals including exactly 2 teachers.

Case 2: Selecting more than 2 teachers
In this case, we need to select 3 or more teachers from the given pool. We can use the complement rule to find the number of ways to select less than 2 teachers and subtract it from the total number of ways to select 6 individuals.

The number of ways to select 0 or 1 teacher is given by:

C(4, 0) + C(4, 1) = 1 + 4 = 5

Therefore, the number of ways to select more than 2 teachers is:

Total ways - Ways to select less than 2 teachers = C(12, 6) - 5 = 924 - 5 = 919

Therefore, there are 919 ways to select 6 individuals including more than 2 teachers.

Total ways:
To find the total number of ways to select 6 individuals including at least 2 teachers, we add the number of ways from both cases:

420 + 919 = 1339

Therefore, there are 1339 ways to select 6 individuals from 4 teachers and 8 students such that at least 2 teachers are included.