# [Combinations//Permutations] Item Arrangments

• August 16th 2010, 03:38 AM
Cthul
[Combinations//Permutations] Item Arrangments
Quote:

Johnathan has 18 different trophies on his shelf. Four of the trophies are for racing, nine are for high jumping, five are for cycling.
How many different displays can be made if:
1, Six trophies are chosen?
$nPr(18,6)=13366080$
2, Six trophies are chosen, two from each activity?

I'm not sure how to do the second one, please help.
• August 16th 2010, 03:55 AM
Quote:

Originally Posted by Cthul
How many different displays can be made if:
1, Six trophies are chosen?
$nPr(18,6)=13366080$
2, Six trophies are chosen, two from each activity?

I'm not sure how to do the second one, please help.

3 pairs of trophies are chosen.
1 pair must be racing trophies, so "select" 2 from 4 to find out how many pairs of racing trophies you can have.

Another pair must be high-jump trophies, so "select" 2 from 9.

Another pair must be cycling trophies, so "select" 2 from 5.

You can combine any pair of racing trophies with any pair of high-jump trophies,
hence multiply the selections.
Those 4 can go with any pair of cycling trophies, so multiply again.

Use nCr for the selections.

For your Q1, you've calculated displays with the 6 trophies in different positions.
Is this feasible?
• August 16th 2010, 04:09 AM
Cthul
Er..
I can do Question 1, I just thought I might add it in to let readers know that I know a some parts.

So it's..
$nCr(5,2)\times nCr(9,2)\times nCr(4,2)\times6!$... I got it right. Thanks.