# 3 questions, involving ratio, percentages and probability

• Aug 28th 2010, 06:23 PM
ethanraiden
3 questions, involving ratio, percentages and probability
Hi, I''m new :) I have 3 questiosn that I'm having trouble with, I'll list the question and then the answer that I think is correct, I'm just not sure about them, so any help you guys could give would be appricatied.

1. Assume that you're testing a boss monster in a game. The monster has the chance to drop 12 different items when it's defeated, but will only drop 4 items per defeat. In addition, you can get multiple of the same item, but that counts towards the 4 possible item drops. How many variations are possible? Answer:20738

2. Assume you're testing a spell called Magic Bolt in a game. After a designer makes a change to cast time, you noticed a 60% increase in damage per second against a normal target. Before the designer changed the spell, you knew it took 2 seconds to cast and did 800 points of damage per cast, at a normal target that would reduce the damage taken from Magic Bolt by a fourth. What percent reduction was applied to the 2 second cast time to create the increase in damage? Answer 50% (really unsure)

3. Assume that you're testing falling damage in a game. From a standing still position, you know that falling from 80 yards above the ground takes exactly 4 seconds to reach the ground and will do 2000 points of damage to you. Also, you know that falling from 500 yards above the ground takes exactly 10 seconds and will do 5000 points of damage. How much damage will your character take at 180 yards above the ground? Answer 2750

Thanks a lot for any help that you might be able to give, I think the most confusing parts of these questions is the wording.
• Aug 28th 2010, 07:09 PM
Prove It
1. If there are 12 different items, at any defeat you will pick up 4 items and you can pick up up to 4 of the same item, that means that there must be $\displaystyle 12 \times 4 = 48$ items to choose from.

So what you will be working out here is the number of different combinations of 4 you can have with 48 items.

This is $\displaystyle {48\choose{4}} = \frac{48!}{4!(48 - 4)!}$

$\displaystyle = \frac{48\cdot 47 \cdot 46 \cdot 45 \cdot 44!}{4!44!}$

$\displaystyle = \frac{48\cdot 47 \cdot 46 \cdot 45}{4\cdot 3\cdot 2\cdot 1}$

$\displaystyle = 2\cdot 47\cdot 46 \cdot 45$

$\displaystyle = 194580$.