A hockey game is preparing for a group photo. The team has 1 goalie, 4 defense, and 7 forwards. The photographer wants two rows of six players. How many ways can the team arrange six players in the front row with at most 2 defense?

case1: no defence in 1st row

- we can choose any 6 of the 8 players
= C(8,6) = 28

case2: 1 defence
- four ways to choose the defenseman, choose 5 from the remaining 8
= 4 x C( 8,5) = = 224
Case3 : 2 defense
= C(4,2) x C(8,4) = 420

total = 28+224+420 = 672 for choosing the 6

but you wanted the number of ways to arrange them ...
number of ways = 672 x 6! = 483,840

You are literally a life saver, thank you so so so so so so much!!

To determine the number of ways the team can arrange six players in the front row with at most 2 defense, we will consider two cases:

Case 1: No defense on the front row:
In this case, all six players in the front row must be either forwards or the goalie. Since there are 7 forwards and 1 goalie, the number of ways to select 6 players from this group is given by the combination formula:

C(7+1,6) = C(8,6) = 8! / (6! * (8-6)!) = 28

Case 2: One defense on the front row:
In this case, there will be 5 players in the front row who are either forwards or the goalie. We also have to consider the position of the defense on the front row.

The number of ways to select 5 players from the forwards and goalie is given by:

C(7+1,5) = C(8,5) = 8! / (5! * (8-5)!) = 56

For the position of the defense, we have 4 defense players available. Since there are 5 positions in the front row, we select 1 position for the defense from the available 5 positions:

C(5,1) = 5

Total number of arrangements for this case = 56 * 5 = 280.

Case 3: Two defense on the front row:
In this case, there will be 4 players in the front row who are either forwards or the goalie. We also have to consider the positions of the 2 defense on the front row.

The number of ways to select 4 players from the forwards and goalie is given by:

C(7+1,4) = C(8,4) = 8! / (4! * (8-4)!) = 70

For the positions of the 2 defense, we have 5 positions available in the front row. We select 2 positions for the defense from the available 5 positions:

C(5,2) = 10

Total number of arrangements for this case = 70 * 10 = 700.

Therefore, the total number of ways the team can arrange six players in the front row with at most 2 defense is given by the sum of the arrangements from all cases:

28 + 280 + 700 = 1008.

Therefore, there are 1008 ways the team can arrange six players in the front row with at most 2 defense.

To determine the number of ways the team can arrange six players in the front row with at most 2 defense, we need to consider a few possibilities.

First, let's consider the scenario where all six players in the front row are forwards. In this case, we need to choose six forwards from the seven available. The number of ways to do this is given by the combination formula: C(7, 6) = 7.

Next, let's consider the scenario where there is exactly one defenseman in the front row. We have two possible cases: 1 defenseman and 5 forwards or 1 defenseman and 1 goalie.

For the first case, we need to choose one defenseman from the four available and five forwards from the seven available. The number of ways to do this is given by: C(4, 1) * C(7, 5) = 4 * 21 = 84.

For the second case, we need to choose one defenseman from the four available and one goalie from the one available. The number of ways to do this is simply: C(4, 1) * C(1, 1) = 4.

Finally, let's consider the scenario where there are two defensemen in the front row. In this case, we need to choose two defensemen from the four available. The number of ways to do this is given by: C(4, 2) = 6.

To find the total number of ways to arrange six players in the front row with at most 2 defense, we sum up the possibilities: 7 + 84 + 4 + 6 = 101.

Therefore, there are 101 ways the team can arrange six players in the front row with at most 2 defense.