| 31 DIFFERENT LINE-UPS?! 20:21 - Dec 30 with 1295 views | Chris_ITFC |
They always talk about a poor manager "not knowing his best team". PL's had 31 attempts, and NEVER gone back to one.
Love him or hate him, it's utterly ridiculous. |  | | |  |
| 31 DIFFERENT LINE-UPS?! on 20:40 - Dec 30 with 1230 views | itfcjoe |
It’s a pretty incredible stat, has it been verified? |  |
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| 31 DIFFERENT LINE-UPS?! on 20:43 - Dec 30 with 1225 views | BLUEBEAT |
That’ll be coming to an end if he actually listened to what the players had to say after the Lincoln match. |  |
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| 31 DIFFERENT LINE-UPS?! on 20:47 - Dec 30 with 1219 views | GeoffSentence |
| 31 DIFFERENT LINE-UPS?! on 20:40 - Dec 30 by itfcjoe |
It’s a pretty incredible stat, has it been verified? |
Unique starting line ups by Bluefish 30 Dec 2019 10:01Geoff Sentence has posted it a lot on here but I couldn't fully believe it so i have just copied and pasted and sorted the line ups. He is spot on, we have had a different line up in all 31 games this season. We aren't talking different to the game before we are talking a line up that has never been used before at any stage of the season. This is absolutely staggering! Surely a club needs a strongest 11 that gets tinkered with slightly due to form and fitness. The most frustrating part is this is off the back of a free hit for 6 months last season where we had time to prepare.
I honestly don't think this would have ever been done before in football
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| 31 DIFFERENT LINE-UPS?! on 21:03 - Dec 30 with 1176 views | Trequartista |
| 31 DIFFERENT LINE-UPS?! on 20:40 - Dec 30 by itfcjoe |
It’s a pretty incredible stat, has it been verified? |
It is true. If you remove the goalkeeper, we have not fielded the same 10 outfield players either. |  |
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| 31 DIFFERENT LINE-UPS?! on 21:48 - Dec 30 with 1102 views | bluejacko |
Surely that has to be some kind of record! Can we put that in the trophy cabinet? |  | | |
| 31 DIFFERENT LINE-UPS?! on 21:51 - Dec 30 with 1084 views | JoHnNnY |
| 31 DIFFERENT LINE-UPS?! on 21:48 - Dec 30 by bluejacko |
Surely that has to be some kind of record! Can we put that in the trophy cabinet? |
That would be a kind of trophy Norwich would want to fill some space. |  |
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| 31 DIFFERENT LINE-UPS?! on 21:58 - Dec 30 with 1070 views | Nthsuffolkblue |
To keep this stat up we must soon be able to predict a line-up unless he brings in 5+ players in the window.
Maybe tomorrow will be:
Holy
Garbutt Wilson Chamber Woolfenden
Sears Skuse Downes Nolan
Norwood Keane
That would certainly be a new one. |  |
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| 31 DIFFERENT LINE-UPS?! on 22:44 - Dec 30 with 1021 views | Chris_ITFC |
| 31 DIFFERENT LINE-UPS?! on 21:58 - Dec 30 by Nthsuffolkblue |
To keep this stat up we must soon be able to predict a line-up unless he brings in 5+ players in the window.
Maybe tomorrow will be:
Holy
Garbutt Wilson Chamber Woolfenden
Sears Skuse Downes Nolan
Norwood Keane
That would certainly be a new one. |
On the basis that he's doing it on purpose? | | | |
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| 31 DIFFERENT LINE-UPS?! on 23:03 - Dec 30 with 999 views | Swansea_Blue |
Every game different. Bonkers. [Post edited 31 Dec 2019 0:32]
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| 31 DIFFERENT LINE-UPS?! on 02:18 - Dec 31 with 935 views | ClareBlue |
OK, so I have sussed it...
PL is a genius
Permutations
The most typical concepts of permutations where arrangements of a fixed number of elements r, are taken from a given set n. Essentially this can be referred to as r-permutations of n or partial permutations, denoted as nPr, nPr, P(n,r), or P(n,r) among others. In the case of permutations without replacement, all possible ways that elements in a set can be listed in a particular order are considered, but the number of choices reduces each time an element is chosen, rather than a case such as the "combination" lock, where a value can occur multiple times, such as 3-3-3. For example, in trying to determine the number of ways that a team captain and goal keeper of a soccer team can be picked from a team consisting of 11 members, the team captain and the goal keeper cannot be the same person, and once chosen, must be removed from the set. The letters A through K will represent the 11 different members of the team:
A B C D E F G H I J K 11 members; A is chosen as captain
B C D E F G H I J K 10 members; B is chosen as keeper
As can be seen, the first choice was for A to be captain out of the 11 initial members, but since A cannot be the team captain as well as the goal keeper, A was removed from the set before the second choice of the goal keeper B could be made. The total possibilities if every single member of the team's position were specified would be 11 × 10 × 9 × 8 × 7 × ... × 2 × 1, or 11 factorial, written as 11!. However, since only the team captain and goal keeper being chosen was important in this case, only the first two choices, 11 × 10 = 110 are relevant. As such, the equation for calculating permutations removes the rest of the elements, 9 × 8 × 7 × ... × 2 × 1, or 9!. Thus, the generalized equation for a permutation can be written as:
nPr =
n!
(n - r)!
Or in this case specifically:
11P2 =
11!
(11 - 2)!
=
11!
9!
= 11 × 10 = 110
Again, the calculator provided does not calculate permutations with replacement, but for the curious, the equation is provided below:
nPr = nr
Combinations
Combinations are related to permutations in that they are essentially permutations where all the redundancies are removed (as will be described below), since order in a combination is not important. Combinations, like permutations, are denoted in various ways including nCr, nCr, C(n,r), or C(n,r), or most commonly as simply ( n )
r
. As with permutations, the calculator provided only considers the case of combinations without replacement, and the case of combinations with replacement will not be discussed. Using the example of a soccer team again, find the number of ways to choose 2 strikers from a team of 11. Unlike the case given in the permutation example, where the captain was chosen first, then the goal keeper, the order in which the strikers are chosen does not matter, since they will both be strikers. Referring again to the soccer team as the letters A through K, it does not matter whether A and then B or B and then A are chosen to be strikers in those respective orders, only that they are chosen. The possible number of arrangements for all n people, is simply n!, as described in the permutations section. To determine the number of combinations, it is necessary to remove the redundancies from the total number of permutations (110 from the previous example in the permutations section) by dividing the redundancies, which in this case is 2!. Again, this is because order no longer matters, so the permutation equation needs to be reduced by the number of ways the players can be chosen, A then B or B then A, 2, or 2!. This yields the generalized equation for a combination as that for a permutation divided by the number of redundancies, and is typically known as the binomial coefficient:
You have got to hand it to him he knows what he is doing... | | | |
| 31 DIFFERENT LINE-UPS?! on 02:35 - Dec 31 with 924 views | ClareBlue |
based on above we win 2.0 on the 1st..happy days (proof will be in the eating of the xmas puding) | | | |
| 31 DIFFERENT LINE-UPS?! on 08:12 - Dec 31 with 829 views | bluejacko |
| 31 DIFFERENT LINE-UPS?! on 02:18 - Dec 31 by ClareBlue |
OK, so I have sussed it...
PL is a genius
Permutations
The most typical concepts of permutations where arrangements of a fixed number of elements r, are taken from a given set n. Essentially this can be referred to as r-permutations of n or partial permutations, denoted as nPr, nPr, P(n,r), or P(n,r) among others. In the case of permutations without replacement, all possible ways that elements in a set can be listed in a particular order are considered, but the number of choices reduces each time an element is chosen, rather than a case such as the "combination" lock, where a value can occur multiple times, such as 3-3-3. For example, in trying to determine the number of ways that a team captain and goal keeper of a soccer team can be picked from a team consisting of 11 members, the team captain and the goal keeper cannot be the same person, and once chosen, must be removed from the set. The letters A through K will represent the 11 different members of the team:
A B C D E F G H I J K 11 members; A is chosen as captain
B C D E F G H I J K 10 members; B is chosen as keeper
As can be seen, the first choice was for A to be captain out of the 11 initial members, but since A cannot be the team captain as well as the goal keeper, A was removed from the set before the second choice of the goal keeper B could be made. The total possibilities if every single member of the team's position were specified would be 11 × 10 × 9 × 8 × 7 × ... × 2 × 1, or 11 factorial, written as 11!. However, since only the team captain and goal keeper being chosen was important in this case, only the first two choices, 11 × 10 = 110 are relevant. As such, the equation for calculating permutations removes the rest of the elements, 9 × 8 × 7 × ... × 2 × 1, or 9!. Thus, the generalized equation for a permutation can be written as:
nPr =
n!
(n - r)!
Or in this case specifically:
11P2 =
11!
(11 - 2)!
=
11!
9!
= 11 × 10 = 110
Again, the calculator provided does not calculate permutations with replacement, but for the curious, the equation is provided below:
nPr = nr
Combinations
Combinations are related to permutations in that they are essentially permutations where all the redundancies are removed (as will be described below), since order in a combination is not important. Combinations, like permutations, are denoted in various ways including nCr, nCr, C(n,r), or C(n,r), or most commonly as simply ( n )
r
. As with permutations, the calculator provided only considers the case of combinations without replacement, and the case of combinations with replacement will not be discussed. Using the example of a soccer team again, find the number of ways to choose 2 strikers from a team of 11. Unlike the case given in the permutation example, where the captain was chosen first, then the goal keeper, the order in which the strikers are chosen does not matter, since they will both be strikers. Referring again to the soccer team as the letters A through K, it does not matter whether A and then B or B and then A are chosen to be strikers in those respective orders, only that they are chosen. The possible number of arrangements for all n people, is simply n!, as described in the permutations section. To determine the number of combinations, it is necessary to remove the redundancies from the total number of permutations (110 from the previous example in the permutations section) by dividing the redundancies, which in this case is 2!. Again, this is because order no longer matters, so the permutation equation needs to be reduced by the number of ways the players can be chosen, A then B or B then A, 2, or 2!. This yields the generalized equation for a combination as that for a permutation divided by the number of redundancies, and is typically known as the binomial coefficient:
You have got to hand it to him he knows what he is doing... |
Mmmm that has got to make sense, problem solved 😀👌 | | | |
| 31 DIFFERENT LINE-UPS?! on 09:21 - Dec 31 with 786 views | hampstead_blue |
| 31 DIFFERENT LINE-UPS?! on 02:18 - Dec 31 by ClareBlue |
OK, so I have sussed it...
PL is a genius
Permutations
The most typical concepts of permutations where arrangements of a fixed number of elements r, are taken from a given set n. Essentially this can be referred to as r-permutations of n or partial permutations, denoted as nPr, nPr, P(n,r), or P(n,r) among others. In the case of permutations without replacement, all possible ways that elements in a set can be listed in a particular order are considered, but the number of choices reduces each time an element is chosen, rather than a case such as the "combination" lock, where a value can occur multiple times, such as 3-3-3. For example, in trying to determine the number of ways that a team captain and goal keeper of a soccer team can be picked from a team consisting of 11 members, the team captain and the goal keeper cannot be the same person, and once chosen, must be removed from the set. The letters A through K will represent the 11 different members of the team:
A B C D E F G H I J K 11 members; A is chosen as captain
B C D E F G H I J K 10 members; B is chosen as keeper
As can be seen, the first choice was for A to be captain out of the 11 initial members, but since A cannot be the team captain as well as the goal keeper, A was removed from the set before the second choice of the goal keeper B could be made. The total possibilities if every single member of the team's position were specified would be 11 × 10 × 9 × 8 × 7 × ... × 2 × 1, or 11 factorial, written as 11!. However, since only the team captain and goal keeper being chosen was important in this case, only the first two choices, 11 × 10 = 110 are relevant. As such, the equation for calculating permutations removes the rest of the elements, 9 × 8 × 7 × ... × 2 × 1, or 9!. Thus, the generalized equation for a permutation can be written as:
nPr =
n!
(n - r)!
Or in this case specifically:
11P2 =
11!
(11 - 2)!
=
11!
9!
= 11 × 10 = 110
Again, the calculator provided does not calculate permutations with replacement, but for the curious, the equation is provided below:
nPr = nr
Combinations
Combinations are related to permutations in that they are essentially permutations where all the redundancies are removed (as will be described below), since order in a combination is not important. Combinations, like permutations, are denoted in various ways including nCr, nCr, C(n,r), or C(n,r), or most commonly as simply ( n )
r
. As with permutations, the calculator provided only considers the case of combinations without replacement, and the case of combinations with replacement will not be discussed. Using the example of a soccer team again, find the number of ways to choose 2 strikers from a team of 11. Unlike the case given in the permutation example, where the captain was chosen first, then the goal keeper, the order in which the strikers are chosen does not matter, since they will both be strikers. Referring again to the soccer team as the letters A through K, it does not matter whether A and then B or B and then A are chosen to be strikers in those respective orders, only that they are chosen. The possible number of arrangements for all n people, is simply n!, as described in the permutations section. To determine the number of combinations, it is necessary to remove the redundancies from the total number of permutations (110 from the previous example in the permutations section) by dividing the redundancies, which in this case is 2!. Again, this is because order no longer matters, so the permutation equation needs to be reduced by the number of ways the players can be chosen, A then B or B then A, 2, or 2!. This yields the generalized equation for a combination as that for a permutation divided by the number of redundancies, and is typically known as the binomial coefficient:
You have got to hand it to him he knows what he is doing... |
Love it.
I think PL is trying to prove a point here.
He has someone sat next to a typewriter, blindfolded, randomly pressing keys.
Eventually he hopes he will come-up with a formation and combination which works.
Only another 3 million years until a thread appears..... |  |
| Assumption is to make an ass out of you and me.
Those who assume they know you, when they don't are just guessing.
Those who assume and insist they know are daft and in denial.
Those who assume, insist, and deny the truth are plain stupid.
Those who assume, insist, deny the truth and tell YOU they know you (when they don't) have an IQ in the range of 35-49.
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| 31 DIFFERENT LINE-UPS?! on 09:25 - Dec 31 with 774 views | GeoffSentence |
| 31 DIFFERENT LINE-UPS?! on 09:21 - Dec 31 by hampstead_blue |
Love it.
I think PL is trying to prove a point here.
He has someone sat next to a typewriter, blindfolded, randomly pressing keys.
Eventually he hopes he will come-up with a formation and combination which works.
Only another 3 million years until a thread appears..... |
Give an infinite number of monkeys and infinite number of typewriters and in time they will get a Paul Lambert team selection right. | |
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| 31 DIFFERENT LINE-UPS?! on 09:33 - Dec 31 with 762 views | Herbivore |
| 31 DIFFERENT LINE-UPS?! on 09:25 - Dec 31 by GeoffSentence |
Give an infinite number of monkeys and infinite number of typewriters and in time they will get a Paul Lambert team selection right. |
I've number crunched and it would take them 31 million years. |  |
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