Eight gymnastics players numbered 1 through 8 underwent a training camp where they were coached by three coaches- Xena, Yuki, and Zara. Each coach trained at least two players. Yuki trained only even-numbered players, while Zara trained only odd-numbered players. After the camp, the coaches evaluated the players and gave integer ratings to the respective players trained by them on a scale of 1 to 7, with 1 being the lowest rating and 7 the highest. The following additional information is known:

1. Xena trained more players than Yuki.

2. Player-1 and Player-4 were trained by the same coach, while the coaches who trained Player-2, Player-3, and Player-5 were all different.

3. Player-5 and Player-7 were trained by the same coach and got the same rating. All other players got a unique rating.

4. The average of the ratings of all the players was 4.

5. Player-2 got the highest rating.

6. The average of the ratings of the players trained by Yuki was twice that of the players trained by Xena and two more than that of the players trained by Zara.

7. Player-4’s rating was double of Player-8’s and less than Player-5’s

For how many players the ratings can be determined with certainty?

The correct option is 6.

Let us solve the puzzle step-by-step to find the ratings of each player and determine how many can be known with certainty.

Step 1: Determine the distribution of players coached by Xena (X), Yuki (Y), and Zara (Z)
We have 8 players, numbered 1 through 8.
Let N(X), N(Y), and N(Z) denote the number of players coached by Xena, Yuki, and Zara respectively.
From the given information:
1. Each coach trained at least 2 players.
2. Yuki (Y) trained only even-numbered players (subsets of {2, 4, 6, 8}).
3. Zara (Z) trained only odd-numbered players (subsets of {1, 3, 5, 7}).
4. Xena (X) trained more players than Yuki, so $N\left(X\right)>N\left(Y\right)$.

Since the total number of players is 8, we have:
$N\left(X\right)+N\left(Y\right)+N\left(Z\right)=8$
Since $N\left(Y\right)\ge 2$ and $N\left(Z\right)\ge 2$, and $N\left(X\right)>N\left(Y\right)$, the only possible distribution is:
$N\left(X\right)=4$, $N\left(Y\right)=2$, and $N\left(Z\right)=2$.
(Any other combination either violates $N\left(X\right)>N\left(Y\right)$ or results in a coach training fewer than 2 players).

Step 2: Assign players to coaches
- Player-1 and Player-4 were trained by the same coach. Since Player-1 is odd (cannot be Y) and Player-4 is even (cannot be Z), their common coach must be Xena (X).
- The coaches who trained Player-2, Player-3, and Player-5 were all different. Since Y only trains even players, Y cannot train 3 or 5. Thus, Player-2 must be trained by Yuki (Y).
- Since Player-2's coach is Y, and 3 and 5 are trained by different coaches, one is trained by X and the other by Z.
- Player-5 and Player-7 were trained by the same coach. Since Z must train exactly 2 players (and Z can only train odd-numbered players), Player-5 and Player-7 must be trained by Zara (Z). If they were trained by X, Z would only train at most 1 player (Player-3), which is a contradiction.
- Thus, Player-3 must be trained by Xena (X).
- Currently, Z's players are {5, 7}.
- X's players are {1, 3, 4} plus one of {6, 8}.
- Y's players are {2} plus the other of {6, 8}.

Step 3: Determine the ratings of the players
Let the rating of Player-i be represented by R(i).
- The average rating of all 8 players is 4. Thus, the sum of all ratings is:
$8\times 4=32$
- Player-5 and Player-7 got the same rating, and all other players got unique ratings.
- Since there are 8 players with ratings from 1 to 7, and exactly one rating is repeated (for Player-5 and Player-7), the set of ratings must contain all integers from 1 to 7 exactly once, with the repeated rating r appearing a second time.
- Therefore:
$(1+2+3+4+5+6+7)+r=32$
$28+r=32\Rightarrow r=4$
- This gives:
$R\left(5\right)=4$ and $R\left(7\right)=4$
- Player-2 got the highest rating, so:
$R\left(2\right)=7$

Step 4: Find the average ratings for each coach
- Zara (Z) coached Player-5 and Player-7, who both got a rating of 4. So the average rating of Z's players is:
$A\left(Z\right)=4$
- According to information 6, the average rating of Y's players, A(Y), is 2 more than that of Z's:
$A\left(Y\right)=A\left(Z\right)+2=4+2=6$
- The average rating of X's players, A(X), is half of Y's:
$A\left(X\right)=\frac{A\left(Y\right)}{2}=\frac{6}{2}=3$

Step 5: Solve for the remaining ratings
- Yuki (Y) trained Player-2 and one player from {6, 8}. Let this second player be P_Y.
- The sum of ratings of Y's players is:
$R\left(2\right)+R\left(P_Y\right)=2\times A\left(Y\right)=12$
- Since $R\left(2\right)=7$, we have:
$R\left(P_Y\right)=12-7=5$
- Player-4's rating is double of Player-8's and less than Player-5's ($R\left(5\right)=4$).
- Since $R\left(4\right)<4$ and $R\left(4\right)=2\times R\left(8\right)$, the only possible integer values are:
$R\left(8\right)=1$ and $R\left(4\right)=2$
- Since $R\left(8\right)=1$ (not 5), Player-8 cannot be P_Y.
- Therefore, Player-6 must be coached by Yuki, so:
$R\left(6\right)=5$
- This leaves Player-8 to be coached by Xena (X).
- The sum of ratings of X's players {1, 3, 4, 8} is:
$R\left(1\right)+R\left(3\right)+R\left(4\right)+R\left(8\right)=4\times A\left(X\right)=12$
- Substituting the values we know:
$R\left(1\right)+R\left(3\right)+2+1=12\Rightarrow R\left(1\right)+R\left(3\right)=9$
- The remaining unused ratings are 3 and 6. Thus, one of Player-1 and Player-3 has a rating of 3, and the other has a rating of 6. However, we cannot determine which is which.

Conclusion:
The ratings determined with certainty are:
- Player-2: 7
- Player-4: 2
- Player-5: 4
- Player-6: 5
- Player-7: 4
- Player-8: 1
Thus, the ratings of exactly 6 players can be determined with certainty.