5. (5 marks) How many ways are there to choose 10 co-op candidates out of 25 students to send to work 
at the same company if four of the students insist on being together (you can select all four of them 
or none of them) and another two students won’t work together (you can select none of them or 
either one of them, but not both of them)? Hint: make a case-by-case argument, for example, one 
case is: employ all 4 friends and one of the enemies and 5 regulars …

This problem can be solved using the method of cases, as suggested in the hint. The strategy involves breaking down the possibilities based on the behavior of two special groups: the **four students who insist on being together** (the "friends"), and the **two students who won't work together** (the "enemies"). 

We’ll look at different cases for the friends and enemies and compute how many ways we can choose the remaining students in each case.

### Case 1: All 4 friends are chosen
In this case, we already have 4 students selected (the 4 friends), so we need to choose 6 more from the remaining 21 students. However, we must account for the constraint that **we can select either 0 or 1 of the enemies, but not both**.

#### Sub-case 1.1: Choose neither enemy
If we select none of the enemies, we must choose 6 regular students from the 19 remaining students. This can be done in:
\[
\binom{19}{6} \text{ ways.}
\]

#### Sub-case 1.2: Choose exactly one enemy
If we select exactly 1 enemy, we must choose 5 more regular students from the remaining 19 students. Since there are 2 enemies, we can select one of them in \(\binom{2}{1} = 2\) ways, and then choose 5 more students from the remaining 19 students. This can be done in:
\[
2 \times \binom{19}{5} \text{ ways.}
\]

So, the total number of ways in **Case 1** is:
\[
\binom{19}{6} + 2 \

How many ways are there to choose 10 co-op candidates out of 25 students to send to work at the same company if four of the students insist on being together (you can select all four of them or none of them) and another two students won’t work together (you can select none of them or either one of them, but not both of them)?

Explore how to solve a combinatorics problem involving 25 students, 4 insisting on being together, and 2 who won’t work together. Learn the case-by-case analysis approach and apply the binomial coefficient formula to determine the number of ways to choose 10 candidates.

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