Re: Statistics Combination Question

Quote:

Originally Posted by

**marcusman** 15 People have been given complementary tickets to 3 shows. [/FONT][/COLOR]

There are:

- 6 tickets for show 1,

- 5 for show 2 and

- 4 for show 3.

How many ways can the group of 15 be divided between the 3 shows if 2 people refuse to go to show 1 and one person insists on going to show 3?

Also, if the tickets end up getting randomly distributed, what is the chance that the 3 people get there wish (i.e. persons 1 and 2 don't go to show 1 and person 3 goes to show 3)

First there are ways to give the tickets out without any restrictions whatsoever.

Now think this way. Give person three a ticket to show three.

So we have reduced the problem.

There are ways to give those remaining tickets out without any restrictions whatsoever.

But we must remove from that number the cases in which person one **or** person two got a ticket not wanted. Use inclusion/exclusion:

.

That is the number of ways in which at least one of the two got a ticket not wanted.

Re: Statistics Combination Question

Hey thanks for the reply. I get all of it except why you minus the 12! equation from the 13! equation

Re: Statistics Combination Question