Embarassingly Basic Maths Question

poshniallo

Hi

This is not homework - it's just maths for a hobbygame I'm thinking about.

I have 3 groups of 5 elements - let's call them A B C D E, M N O P Q, and V W X Y Z.

If a given entity is assigned one from each group, e.g. AMV or ANW, or even CYV - how many possible entities are there?

This is assuming, by the way, that AMV and VMA are distinct from each other. Order matters.

I think it's a huge amount - 625 maybe? But I'm not sure.

Can someone help me out please? Thanks.

Richy

poshniallo wrote: »

If a given entity is assigned one from each group, e.g. AMV or ANW, or even CYV

Y and V are in the same group. Is this a typo or do you allow multiple elements from the same group?

BSoB

5*5*5 is order doesn't matter

order does matter adds a *3! AKA 6

so 5*5*5*6 is 750?

its been awhile since i've done prob, but that makes my sanity check.

Richy

If you only allow one element per group:
You have five elements in each group, so the number of possible combinations is 5*5*5. Then each combination can be written in 6 different ways by changing the order. So the total is 5*5*5*6 = 750.

Bowen

Do the numbers get used up?

15*14*13 = 2730 combinations ?

I have terrible math when it comes to non-algebra stuff though.

Also, CYV doesn't come from each group.

McKid

You already said that order doesn't matter, but we need to know :
1)Can you pick more than an element in each group ?
2)If yes, can the same element be picked more than once ?

poshniallo

Order does matter, and that was a typo about Y and V being in the same group.

Thanks! That makes complete sense. I knew it was about 5 cubed, and I knew it was about 3!, but I couldn't remember what to do with them.