AbdurRakib wrote:

A group of students, each of whom has prepared a presentation, is asked to choose who will give presentations during the next class meeting and in what order they will be given. How many different ways are there to select and arrange the order of presentations?

(1) Five students must be chosen.

(2) All but three students must give presentations at the next class meeting.

Dear

AbdurRakib,

I'm happy to respond.

This is a good question.

Statement #1:

Five students must be chosenHere, we don't know the pool. If there are only five students total, then there is only one group of individuals, and it is simply a matter of arranging them. If there are 100 students, then there would be a mindbogglingly large number of combinations of the five people that could be picked (you know have to know how to calculate this, but it's actually over seventy-five million!!) The number depends in part on the size of the pool. This statement, alone and by itself, is

insufficient.

Now, we have to be very careful to blot the first statement from our minds and consider statement #2 in its pristine solitude.

Statement #2:

All but three students must give presentations at the next class meeting.

Now, we know the number of "not chosens," but we have no idea how many are chosen. It could be that there are four students total, and one gives the presentation in the next class: this would result in simply four ways of choosing. Or there could be six students, and three give it next class: for this, even with out a calculation, we can see that there would be more than four ways of choosing. Without knowing definitely the number chosen, we cannot answer the prompt question. This statement, alone and by itself, is

insufficient.

Now, consider the two statement together.

Now, five are chosen for next class, and 3 remain unchosen, so the total number of students with presentations is 8. How many combinations of 5 students can we pick from the 8? That's 8C5. Once, we have those five, we can put them in 5! orders. Thus, the total number of ways is

N = (8C5)(5!)

We don't need to calculate. It's enough to know that we could. Together, both statements are

sufficient.

OA =

(C) For those curious about the calculation:

8C5 = 56

5! = 120

N = (8C5)(5!) = (56)(120) = 56(100 + 20) = 5600 + 1120 = 6720

Mike

_________________

Mike McGarry

Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)