Combinations?

Question

I am stuck on this question, and i really really need help! please please!

A project team of 6 students is to be selected from a class of 30.... Pierre, Gregory, and Miguel are students in this class. How many of the teams would include these 3 students...

I know the answer is 2925, but i have no idea how to get that answer!

Thanks thanks thanks!

Answer 1

27 choose 3.

Essentially, take those 3 kids out of the group of 30 and put them in the selected group of 6.

Now, you have 27 kids left, of which you need to select 3 to form the 6 group committee.

27 choose 3 = 2925.

Answer 2

You know 3 members of the team are Pierre, Gregory, and Miguel, so you're basically trying to find the number of ways to choose the other 3 students for the 6-person team.

There are 27 students left in the class, and 3 spots to be filled. Since the order doesn't matter, you use a combination (rather than a permutation). You would do the calculation "27-choose-3", or

com[27,3]

which equals 2925.

Answer 3

If there are no restrictions on the teams, the solution is the number of combinations of 30 items taken 6 at a time. However, if three members must be in the team, we are reduced to the number of combinations of 27 items taken 3 at a time or 27!/(3!x 24!)

Answer 4

Combinations: There are many problems in which we are interested in determining the number of ways in which k objects can be selected from n distinct objects without regard to the order in which they are selected. Such selections are called combinations or k-sets. It may help to think of combinations as a committee. The key here is without regard for order.

The number of combinations of k objects from a set with n objects is n C k. For example, the combinations of {1,2,3,4} taken k=2 at a time are {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, for a total of 6 = 4! / [(2!)(4-2) !] subsets.

The general formula is:


n C k = n! / [k! (n-k) !].

now in your case n = 27 because you know that each team must contain Pierre, Greroy, and Miguel

k = 3 (unknown of teams are 3)

27!/(3!*(27-3)!) = 27!/(3!*24!) = 27*26*25/(3*2) = 2925

the website below explains this

Answer 5

27C3= 2925
(Since the three must be included, effectively, you will be choosing 3 among the remaining 27.)