a total of 77 football players and 11 starting positons

If there are a total 77 football players on a team, and 11 positions, how many times would it take until every player on the team can say they've played with every other player at least once? (I hope that is clear.)

This sounds like homework. We don't do homework...at least not without seeing you own attempts first.

Also, define "time."

That said, I would say the situation either never occurs (because offense and defense never play together), or it can occur within the first "time", with "time" being a game. With current rules regarding substitution, every player should be able to get into at least one play, so, within one game, they have all played together.

I enjoy your interesting questions Ed and it looks like I am not the only one. The Wizard has responded to your last 2 questions so maybe he will chime in on this one as well.

Agreeing with DJ - the question needs a lot more clarifications to even be able to attempt to answer.

Quote: DJTeddyBear

This sounds like homework. We don't do homework...at least not without seeing you own attempts first.

Also, define "time."

That said, I would say the situation either never occurs (because offense and defense never play together), or it can occur within the first "time", with "time" being a game. With current rules regarding substitution, every player should be able to get into at least one play, so, within one game, they have all played together.

I'd assume for "football", you crazy Americans should read "soccer" :)

I'm 50 years old and I haven't had any kind of homework in 30+ some odd years. It is certainly not homework coming from me. :)

HOWEVER, this question was asked to me, and so it certainly MIGHT be homework from someone else. :)

After re-reading the question, I agree it is not as clear as it could be. Sorry. Let's see if I can completely rephrase it.

There are a total of 77 players that make up a football team. 11 positions are available in football and thus 11 players can be on the field at any one time. Each player would like to be able to say they were on the field, (and involved in a play) with EACH of the other 76 players. How many different arrangements will this require?

For player #1 to say that, I've come up with eight different arrangements.

1 2 3 4 5 6 7 8 9 10 11
1 12 13 14 15 16 17 18 19 20 21
1 22 23 24 25 26 27 28 29 30 31
1 32 33 34 35 36 37 38 39 40 41
1 42 43 44 45 46 47 48 49 50 51
1 52 53 54 55 56 57 58 59 60 61
1 62 63 64 65 66 67 68 69 70 71
1 72 73 74 75 76 77 2 12 13 14

As shown above, if this were the 11 players who were on the field for the first eight plays of the game, Player #1 could later state that he was involved in a play with each of the other 76 players on his team.

My computer programming skills are much better than my math/probability/permutation/combination skills. In order for me to solve this, I would have to write a program to loop through all the different permutations, and counting the number of arrangements were the criteria is met. But I'm pretty sure this problem is easily solved with a very simple permutation/combination formula.

Quote: DJTeddyBear

Also, define "time."

The quality of existence that prevents everything from taking place simultaneously :P

Quote: Nareed

Quote: DJTeddyBear

Also, define "time."

The quality of existence that prevents everything from taking place simultaneously :P

Well, I had meant in the context of the question being posed...

But if you insist, here's how Wikipedia defines time. Or the first sentence anyway:

Quote: Wikipedia

Time is the indefinite continued progress of existence and events that occur in apparently irreversible succession from the past through the present to the future.

I love how they felt the need to put 'apparently' in there. LOL.

Quote: thecesspit

I'd assume for "football", you crazy Americans should read "soccer" :)

HEY! I resemble that remark!

Ya gotta understand that this is website is owned by an American, with a primary focus upon a city in America, and with a mostly American membership.



For the record, I really have no idea how American Football got it's name. After all, there's hardly any kicking involved, and when the kicker DOES score, it's for less points than a touchdown!

Why would there be 77 soccer players on a team? That's why american football makes much more sense.

Quote: DJTeddyBear

Well, I had meant in the context of the question being posed...

Yes, well, that wouldn't have been as funny.

Quote:

But if you insist, here's how Wikipedia defines time. Or the first sentence anyway:I love how they felt the need to put 'apparently' in there. LOL.

Seriously, time is one of those concepts that everyone understands yet few people can define. As for the direction of time, there's an idea I've heard about, apparently from Richard Feynman, that anti-matter is really regular matter moving backwards in time. But that goes way above my pay grade.

Oh, yes, 77 players are a college football team, not a soccer one.

I think I can rephrase the question:

A lottery game consists of each ticket having 11 numbers from 1 to 77. Two numbers (again, from 1 to 77) are drawn, and any ticket with both numbers wins. What is the minimum number of tickets needed to guarantee that at least one ticket will win, regardless of the numbers drawn?

I have read about problems like this, but I don't know if there's a general solution.
Here is a description of the general problem.

It sounded vaguely like the lottery wheel problem when I read it too. There's a trivial lower bound:

a) There are 77C2 = 2926 combinations of two numbers from 77;
b) Each ticket (or team) has 11C2 = 55 combinations
c) 2926/55 = 53.2, so 54 tickets (or team arrangements) is a lower bound. This assumes near-zero repetition of any given pair of numbers.

However, the OP's example on the first page shows 4 repetitions in the first 8 combinations, so I doubt 54 is achievable.

An alternate statement of the problem is this:

Given a complete undirected graph G = (V, E), V = {1 .. 77}. Since G is complete, E = all possible edges over V.
Let G'n be a complete undirected graph over V, E, where each G'n contains exactly 11 vertices in V (and, as above, all the edges).
Find the smallest N such that the union of G'1 .. G'N = G.

I found a paper:

Lottery paper

Quote: DJTeddyBear

HEY! I resemble that remark!

Ya gotta understand that this is website is owned by an American, with a primary focus upon a city in America, and with a mostly American membership


For the record, I really have no idea how American Football got it's name. After all, there's hardly any kicking involved, and when the kicker DOES score, it's for less points than a touchdown!

Absolutely, I understand that :) I thought I was helping with a little translation.

Football came from a kicking the ball on the ground game, to a en masse melee with a round ball and back again. Sheffield Rules football in the 1850's allowed the handling of the ball if caught directly from a kick, and obviously we also have rugby football where the ball is kicked every so often. Cambridge Rules Football also had it's rules about kicking the shins of a ball carrier. In the 1870's football (aka soccer) pretty much eliminated the handling of the ball apart from the throw-in and the goal keeper.

Harvard and McGill had their own variations on the game, which was much more based on Rugby Football, and these rules became popular in North America. Rugby Football was originally much more like the en-masse game, with the ball being something shoved and fought over on the ground.

For the record, the American version of the game is my favourite (though it has it roots in the McGill Football rules... American Football came via Canada... just like your basketball did :p)

Quote: EdCollins

For player #1 to say that, I've come up with eight different arrangements.

1 2 3 4 5 6 7 8 9 10 11
1 12 13 14 15 16 17 18 19 20 21
1 22 23 24 25 26 27 28 29 30 31
1 32 33 34 35 36 37 38 39 40 41
1 42 43 44 45 46 47 48 49 50 51
1 52 53 54 55 56 57 58 59 60 61
1 62 63 64 65 66 67 68 69 70 71
1 72 73 74 75 76 77 2 12 13 14

As shown above, if this were the 11 players who were on the field for the first eight plays of the game, Player #1 could later state that he was involved in a play with each of the other 76 players on his team.

This seems like the most straightforward way to do it, while quite time consuming. Let me try to finish.

2, 15-24
2, 25-34
2, 35-44
2, 45-54
2, 55-64
2, 65-74
2, 75,76,77,3,12,13,14,15,16,17

3, 18-27
3, 28-37
3, 38-47
3, 48-57
3, 58-67
3, 68-77

Okay, so I'm already a little worn out already, so I'm going to make a generalization and hope it's right. 1 had 8 combos, 2 had 7 combos, 3 had 6 combos. 4 will have 5, 5 will have 4, 6 will have 3, 7 will have 2, 8 will have 1. Maybe. Possibly this covers everything?? ... thinking about I think probably not, so it's probably not just 36 unique combos.

Time is the measure of change.

Quote: MathExtremist

I found a paper:

Lottery paper

That paper is quite comprehensive. After skimming it, I'm convinced the answer to the OP's question is ridiculously hard to determine.

OP wants the value of L( 77 , 11 , 2 ; 2 ). Let's see if they have good bounds for arbitrary values...