Problem Solving

An oratorical society consists of six members, and at an upcoming meeting, the members will present a total of four speeches. If no member presents more than two of the four speeches, in how many different orders could the members give speeches?



(A)720



(B)1080



(C)1170



(D)1470



(E)1560

First of all, this is a very hard/time-consuming question, about as complex as you'd ever see on a test, and perhaps even beyond that. It's good to understand it, but don't get too worried about it.

If four different people give speeches:
6 * 5 * 4 * 3 = 360 possibilities.

If two people each give two speeches:
There are 6C2 possible sets of 2 people, since there are 6 people to choose from
6C2 = 6!/(4!2!) = 6*5/2 = 15

For each set of two people, there are 4C2 orders in which they could give the speeches, since there are 4 speech slots and each gives 2
4C2 = 4!/(2!2!) = 4*3/2 = 6

15*6 = 90 ways to arrange 2 people giving two speeches

If two people give 1 speech each and one person gives 2 speeches:
There are 6 choices for the person who gives 2 speeches
Then, there are 5C2 choices for the 2 people to give one speech each
5C2 = 5!/(3!2!) = 5*4/2 = 10
So there are 6*10 = 60 ways to arrange 1 person to give 2 speeches and 2 people to give 1

Then, there are 4!/2! ways to arrange the 4 speeches by each of these sets of people, because this is a case of arranging AABC, where A is a "repeated element"
4!/2! = 12
So, there are 60*12 = 720 ways to arrange 1 person giving 2 speeches and 2 people giving 1 speech

Add it up: [spoiler]360 + 90 + 720 = 1170[/spoiler]