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Final Fantasy IV Math

cvxfreak

Member
The given: At the end, we have ten possible party members to use. We can take five at a time, and the other five stay behind. One is always present in the party no matter what.

Question: So, how many different combinations are possible?
 
lol I should be easily able to answer this since I took stats last year, but I was really bad at it... I think the number of permutations for a number n is n! So if you only had 5 members, it would be 5! or 120.

I'm not sure how the 10 effects it.. I would have to look in a text.

The amount of different combinations for 10 is 3628800, but you only are dealing with 5 at a time so it's much less than that.
 
I remember how to solve these problems a year ago but I already forgot the formulas. Wario64 cant remember shit :(

Btw, isn't the # of combinations really for 9 people, since 1 person is always gonna be in the party? And does order not matter?
 
Wario64 said:
I remember how to solve these problems a year ago but I already forgot the formulas. Wario64 cant remember shit :(

Btw, isn't the # of combinations really for 9 people, since 1 person is always gonna be in the party? And does order not matter?

If anything it's out of the 9, how many combinations of four of ANY of them are possible?
 
cvxfreak said:
One is always present in the party no matter what.

What does this mean? The same fixed character is always in every party combo? Let's pick Cecil as the fixed member. So Cecil is in every party combo?

argh, slow GAF lag
 
NotMSRP said:
What does this mean? The same fixed character is always in every party combo? Let's pick Cecil as the fixed member. So Cecil is in every party combo?

That's exactly it, and I put it that way to avoid spoiling. ;)
 
cvxfreak said:
If anything it's out of the 9, how many combinations of four of ANY of them are possible?

Let see....

If order does matter. it's 9 * 8 * 7 * 6 = 3024

However, it is doesn't, it is 9 * 8 * 7 * 6 / (4 * 3 * 2 * 1) = 126
 
Since one always has to be in the party, you are selecting the other four out of a group of nine. Since the order doesn't matter, it's a combination, not a permutation:

nCk = n! / k!(n-k)!

9C4 = 9! / 4!(9-5)! = 9! / 4!5! = (362880)/(120*24) = 126




Edit: Late because I was typing up the general formula...<SIGH>
 
Golden Darkness said:
Let see....

If order does matter. it's 9 * 8 * 7 * 6 = 3024

However, it is doesn't, it is 9 * 8 * 7 * 6 / (4 * 3 * 2 * 1) = 126

Thank the lord order doesn't matter. ;)

126 is a lot more feasible than 3024. Thanks everyone! :D
 
Combination without repetition:

30eef8a6e52f56bd1908c4ab3202b91b.png


9!/4!(9-4)!

9!/4!5!

987654321/4321 * 54321

9876/432

3024/24

126

argh, slow again :(

This area of math in counting is called combinatorics.
 
snap.

In the absence of any other killer RPGs I started this game today for the first time (translated snes version on psp emu) and its terrific! Great music, great art, I'm looking forward to making my way through the story.

Not sure if the math in this thread is gonna be any help though. :P
 
Now if math questions were presented in videogame context like that when I went to school, I might have actually given a damn.

eg.

1. It takes a person 80 hours to finish Final Fantasy VII. They have an 8 percent chance of a character dying in every random battle. A random battle happens on average 16 times an hour. How many Phoenix Downs do they need?

2. A player is attempting a speed-run of Mario 3. If each level takes approximately 85 seconds to finish and each mini-game takes 35 seconds, how long will their speed run take if the player uses no warp whistles?
 
Grug said:
Now if math questions were presented in videogame context like that when I went to school, I might have actually given a damn.
I'm sure games could make some great story problems. If Billy Mitchell eats every ghost possible, what's his Pac-Man score after the first 10 stages?
 
Is this one of those statistics that's made up?
 
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