1. ## Combination Question

Hey guys, I'd appreciate any help on this stats question... working would also be appreciated so I can learn how its done

15 People have been given complementary tickets to 3 shows.

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)

2. ## Re: Statistics Combination Question

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 $\displaystyle \frac{15!}{(6!)(5!)(4!)}$ 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 $\displaystyle \frac{14!}{(6!)(5!)(3!)}$ 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:
$\displaystyle 2\frac{13!}{(5!)(5!)(3!)}-\frac{12!}{(4!)(5!)(3!)}$.
That is the number of ways in which at least one of the two got a ticket not wanted.

3. ## 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

4. ## Re: Statistics Combination Question

Originally Posted by marcusman
Hey thanks for the reply. I get all of it except why you minus the 12! equation from the 13! equation
This all about the inclusion/exclusion principle.

$\displaystyle \|A \|$ stands for the number of elements in set $\displaystyle A$.

Do you know to calculate $\displaystyle \|A\cup B\|~?$
Let $\displaystyle A$ be the cases where A receives an unwanted ticket.
Let $\displaystyle B$ be the cases where B receives an unwanted ticket.