Why would this equation work?

[diabloII]

any help would help...well i'm getting help on monday, but just in case someone knew then i wouldn't have to spend 20 minutes getting help

for example:

--> how many different 2 color arrangements can be drawn from a group of 3 different colored cubes?
-->how many different 3 color arrangements can be drawn from a group of 4 different colored cubes?

The equation for both of them is n! over (n-r)! when n=diff colored cubes and r=color arrangements

i.e. (4!=4*3*2*1=24)

why would this equation work?

thx

 

[Chankama]

Easy. There are n objects and we need to select r of those objects where the "order" matters. For example, if the colors are indicated by the numbers 1-4, and we select 3 of em, the arrangement 1,2,3 is different from 2,3,1.

The number of ways of selecting the first object is n. The second object is (n-1), and the r^{th} object is (n-r+1). Therefore, the total number of ordered groups are n x (n-1) x ... x (n-r+1).

And, n x (n-1) x ... x (n-r+1) = n! / (n-r)!

 

[contry]

yeah.. whatever he said >_>' lol

 

[TheMajor]

I think it is 5 combinations for the first one and 24 for the second one. (3*2 and 4*3*2).

 

[Chankama]

I think it is 5 combinations for the first one and 24 for the second one. (3*2 and 4*3*2).

5=3*2????

Yeah. It is 6 and 24. This is exactly what the simple formula produces as well. n! / (n-r)!

 

[desiboi]

HUH?

 

[TheMajor]

5=3*2????

Yeah. It is 6 and 24. This is exactly what the simple formula produces as well. n! / (n-r)!

My bad. It should be 6.

 

[diabloII]

yay

thx a lot

 

[desiboi]

Diablo, why did you copy my sig?

 

[diabloII]

?

i didnt

it just happened to be the same

but seriously

i didn't see i in your sig