Permutation

[pm]

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

[gmatboost]

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]