How many possible team combinations are there?

Cardmant·Jan 9, 2016·2 votes·6,196 views

I was just wondering out of curiosity, how many possible team combinations are there at the moment? E.G which is 1

8 Comments

IronwallJacksonJan 9, 20163 votes

It's time to put my higher education to work. Here's how the math breaks down:

There are currently 128 champions in League, there are 5 spots to a team, and each champion can theoretically fill every role to varying degrees of success.

For the sake of simplicity, let's say that there are 5 roles, 5 slots, and 5 champions. There are two ways to answer your question; using a permutation or a combination. If we count, say, ap mid Ezreal as being different from AD Carry Ezreal, then that means we're working with a permutation, and our formula is n!/(n-k)!, where n is our champion total (128) and k is the number of roles available on a given team (5).

Going by that, we get 128!/5!(128-5)!, which is equal to 31,747,968,000. There close to 32.748 BILLION possible team comps.

However, if you don't care what a champion's role is in a given team comp, then we're working with a combination, and we have to use n!/r!(n-r)!, where, where our variables are the same, but we input for r instead of k.

If we go with that option, then the answer is 128!/213!, or 264,566,400. There are over 264 million team comps made up of unique champions.

TLDR:

31,747,968,000 team comps if you care about exact roles and 264,566,400 if you dont.

4thHorsmanJan 9, 20161 votes

{quoted}

It's time to put my higher education to work. Here's how the math breaks down:

There are currently 128 champions in League, there are 5 spots to a team, and each champion can theoretically fill every role to varying degrees of success.

For the sake of simplicity, let's say that there are 5 roles, 5 slots, and 5 champions. There are two ways to answer your question; using a permutation or a combination. If we count, say, ap mid Ezreal as being different from AD Carry Ezreal, then that means we're working with a permutation, and our formula is n!/(n-k)!, where n is our champion total (128) and k is the number of roles available on a given team (5).

Going by that, we get 128!/5!(128-5)!, which is equal to 31,747,968,000. There close to 32.748 BILLION possible team comps.

However, if you don't care what a champion's role is in a given team comp, then we're working with a combination, and we have to use n!/r!(n-r)!, where, where our variables are the same, but we input for r instead of k.

If we go with that option, then the answer is 128!/213!, or 264,566,400. There are over 264 million team comps made up of unique champions.

TLDR:

31,747,968,000 team comps if you care about exact roles and 264,566,400 if you dont.

Mine wasn't anything to do with roles though, it was simply the champions into the selection pool. For slot 1 there are 128 champs slot 2 127 slot3 126 slot 4 125 and slot 5 124 To get the total possible outcomes you would have to multiply all these amounts together, similar to slot 1 being length, slot 2 being width, slot 3 being depth, and the other two being unknown dimensions xD

IronwallJacksonJan 9, 20161 votes

In permutations, order matters. In other words, if first and third positions changed with each other and the rest of the order stayed the same, it would still be considered a different arrangement. This is important in order to distinguish between permutations and combinations. (Order does not matter for combinations; "1,2,3" is the same as "3,1,2")

What you used is the shorthand for solving permutations. 128!/5!(128-5)! is the same as 128x127x126x125x124. Here's the thing though, when you solve this particular problem as a permutation, rather than a combination then Aatrox , Ahri , Ashe , Alistar , and Amumu add 120 possibiities to our total instead of just 1.

Now, in order for those results to warrant inclusion instead of defaulting to a combination formula instead, there must me a significant difference between them, right? To that end, I assigned meaning to the presentation order by adding a role to each slot. It doesn't matter which one goes where so long as it's consistent and top, mid, supp, adc, and jg are accounted for.

TLDR: both of us calculated for permutations, which gives you 5 times as many results as combinations do, so I made those extra results meaningful by accounting for role distribution.

4thHorsmanJan 10, 20161 votes

{quoted}

What you used is the shorthand for solving permutations. 128!/5!(128-5)! is the same as 128x127x126x125x124. Here's the thing though, when you solve this particular problem as a permutation, rather than a combination then , , , , and add 120 possibiities to our total instead of just 1.

Now, in order for those results to warrant inclusion instead of defaulting to a combination formula instead, there must me a significant difference between them, right? To that end, I assigned meaning to the presentation order by adding a role to each slot. It doesn't matter which one goes where so long as it's consistent and top, mid, supp, adc, and jg are accounted for.

TLDR: both of us calculated for permutations, which gives you 5 times as many results as combinations do, so I made those extra results meaningful by accounting for role distribution.

Derp, you're right. Teach me to do math while still half asleep xD