Possible 4-4-2 Formations

According to the question, the team goes in with a commom $4-4-2$ formation with a goalkeeper. The team has $6$ defenders, $5$ midfielders, $3$ forwards and $2$ goalkeepers. So, in order to find the number of different groups of starting $11$ , we need to find the number different groups of starting defenders, number different groups of starting midfielders, number different groups of starting forwards and number different groups of starting goalkeepers.Let us find them, Number of different groups of defenders $=^{6}C_{4}$ $=15$ Number of different groups of forwards $=^{3}C_{2}$ $=3$ Number of different groups of midfielders $=^{5}C_{4}$ $=5$ Number of different groups of goalkeepers $=^{2}C_{1}$ $=2$ In order to find the total number of different groups of starting $11$ , we need to multiply the number of groups of the constituents of the team. Thus, the total number of different groups of starting $11$ $=15 \times 3 \times 5 \times 2$ $=\boxed{450}$

We have 6 choose 2 for defenders, resulting in 15 possible combinations. We have 5 choose 4 for midfielders, resulting in 5 possible combinations. We have 3 choose 2 for fowards, resulting in 3 possible combinations. We have 2 choose 1 for goalkeepers, resulting in 2 possible combinations. 15x5x3x2=450.

Since each player can only play his designated position, we can consider the players in any position separately from the other players. So, for example, the team has 6 defenders and needs 4 to start, so there are ${6 \choose 4} = 15$ ways to choose them. We repeat this procedure for the rest of the positions and then multiply the number of ways to choose players altogether, to get the number of different groups that they can possibly start with:

${6 \choose 4} \times {5 \choose 4} \times {3 \choose 2} \times {2 \choose 1} = 15 \times 5 \times 3 \times 2 = \boxed {450}$

Thus, there are $\boxed {450}$ ways to choose the 11 starting players.

Let (a,b) be combination a, b. Then, the answer is simply (6,4) (5,4) (3,2)*(2,1)=450

defenders = 6!/(4! 2!) = 15

midfielders = 5!/(4!1!) = 5

forwards = 3!/(2!1!) = 3

goalkeepers = 2!/(1!1!) = 2

total formation = 15 x 5 x 3 x 2 = 450

For this problem, we first check how many possible combinations in each position. We don't use permutation, ( let the four players have $(1,2,3,4)$ as their jersey number respectively), since when players $(1,2,3,4)$ are chosen, it is the same case as when player $(4,3,2,1)$ are chosen.

1. Number of possible combinations to choose 4 defenders in 6 of them, we have $^6\text{C}_4 = 15$ possible ways.

2. Number of possible combinations to choose 4 midfielders in 5 of them, we have $^5\text{C}_4 = 5$ possible ways.

3. Number of possible combinations to choose 2 forwards in 3 of them, we have $^3\text{C}_2 = 3$ possible ways.

4. Number of possible combinations to choose 1 goalkeeper in 2 of them, we have $^2\text{C}_1 = 2$ possible ways.

Now, we have to multiply them, since the team must have 11 players exactly. So, we have $15\times5\times3\times2 = \boxed{450}$ different groups.

For goalkeeper : 2C1 = 2 For Defenders : 6C4 = 15 For Midfielders : 5C4 = 5 For Forward players : 3C2 = 3 Different groups of 11 starting players : 2x15x5x3 = 450

Find a combination between a total of players in every position and a players in rules in every position

6C4 x 5C4 x 3C2 x 2C1 = 15 x 5 x 3 x 2 = 450

$6C4 * 5C4 * 3C2 * 2C1 = 450$

we have to select 11 players from the given conditions... Since 11 players need to be selected, there is an and operation between them..i.e. 4 defenders need to be selected and 4 mid need to be selected and 2 attackers need to be selected and 1 goalkeeper.. Therefore defenders selected in C(6,4)=15 mid selected in C(5,4)=5 attack selected in C(3,2)=3 goalkeeper selected in C(2,1)=2

Since there is an and operation above answers are multiplied to get the final answer = 450

Let b(x,y) be defined as binomial coefficient x over y. The solution is b(2,1) * b(6,4) * b(5,4) * b(3,2) = 450

different groups of 11 starting players =6C4 x 5C4 x 3C2 x 2C1=450...YNWA