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 $\displaystyle \frac{15!}{(6!)(5!)(4!)}$ 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.

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

Quote:

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.