Question

Suppose you are managing 17 employees, and you need to form three teams to work on different projects. assume that all employees will work on a team, and that each employee has the same quali?cations/skills so that everyone has the same probability of getting choosen. in how many different ways can the teams be chosen so taht the number of employees on each project are as follows: 6, 3, 8

Answer 1

To form the first group, we can choose from a group of 17 people.
$\text{First group: } ^{17}C_6 = 12376$

To form the next group, we need to consider how many people are left (17 - 6 = 11)
So, from a group of 11 people, we can have 3 people in it.
$\text{Second group: } ^{11}C_3 = 165$

There is only one way to group the last 8 people. Thus, they are going to be 1 group.

BUT we would have overcounted because of the fact that they can be chosen in different ways. This is because they are not distinct groups or ordered groups. For example, the third group can be chosen first, the second can be chosen last, etc.

So, to counter this, we need to divide by 3!

$\text{Total arrangements: }\frac{12376 \cdot 165}{3!} = 340340$

There are 340, 340 ways.