Question

Use the fundamental counting principle. The students in the 14​-member advanced communications design class at Center City Community College are submitting a project to a national competition. They must select a 5​-member team to attend the competition. The team must have a team leader and a main​ presenter; the other 3 members have no particularly defined roles. In how many different ways can this team be​ formed?
i would also like for you to explain how to simplify this??? 14i would be on top 3i(11i) i just dont get it.. thank you

Answer 1

Answer:

Team can be formed in 40040 different ways.

Step-by-step explanation:

This is a question where three important concepts are involved: permutations, combinations and the fundamental counting principle or multiplication principle.

One of the most important details in the problem is when it indicates that "[...]The team must have a team leader and a main presenter" and that "the other 3 members have no particularly defined roles".

This is a key factor to solve this problem because it is important the order for two (2) positions (team leader and main presenter), but no at all for the rest three (3) other positions.

By the way, notice that it is also important to take into account that no repetition of a team member is permitted to form the different teams requested in this kind of problem: once a member have been selected, no other team will have this member again.

The fundamental counting principle plays an interesting role here since different choices resulted from those teams will be multiplied by each other, and the result finally obtained.

We can start calculating the first part of the answer as follows:

First Part

How many teams of 2 members (team leader and main presenter) can be formed from 14 students? Here the order in which these teams are formed is crucial. There will be a team leader and a main presenter, no more, formed from 14 students.

This part of the problem can be calculated using permutations:

$\frac{n!}{(n-k)!}$ or $\frac{14!}{(14-2!)}= \frac{14*13*12!}{12!}$.

Since $\frac{12!}{12!}=1$, then the answer is $14*13$.

In other words, there are 14 choices to form a team leader (or a main presenter), and then, there are 13 choices to form the main presenter (or a team leader), and finally there are 14*13 ways to form a 2-member team with a leader and a main presenter from the 14 students available.

Second Part

As can be seen, from the total 14 members, 2 members are out for the next calculation (we have, instead, 12 students). Then, the next question follows: How many 3-member teams could be formed from the rest of the 12 members?

Notice that order here is meaningless, since three members are formed without any denomination, so it would be the same case as when dealing with poker hands: no matter the order of the cards in a hand of them. For example, a hand of two cards in poker would be the same when you get an ace of spades and an ace of hearts or an ace of hearts and an ace of spades.

This part of the problem can be calculated using combinations:

$\frac{n!}{(n-k)!k!}$ or $\frac{12!}{(12-3)!*3!}= \frac{12*11*10*9!}{(9!*3!)}$.

Since $\frac{9!}{9!}=1$, then the anwer is $\frac{12*11*10}{3*2*1} = \frac{12}{3}*\frac{10}{2}*11=4*5*11$.

Final Result

Using the multiplication principle, the last thing to do is multiply both previous results:

How many different ways can the requested team be formed?

14*13*4*5*11 = 40040 ways.

Because of the multiplication principle, the same result will be obtained if we instead start calculating how many 3-member teams could be formed from 14 members (combinations) and then calculating how many 2-member team (team leader and main presenter) could be formed from the rest of the 11 team members (permutations).