Statistics Combination Question

[marcusman]

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?

 

[pka]

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?

Give the person who wants a ticket for show 3 his/her ticket.
Give out the six tickets for show 1 to the other twelve people who will take them.
Give out the 5 tickets for show 2 to the eight people left.
The last three tickets go to the last three people.
\(\displaystyle \binom{12}{6}\binom{8}{5}\).

 

[marcusman]

Hey again, thanks for the help.

i understand it all now, just want to check that my answer it right.

I get 1C1 x 12C6 x 8C5 x 3C3 = 51744

(sorry I don't know how to use the formatting on here).

Is that correct?