Cricket has seen so many changes over a period of time. A new concept 2 vs. 2 is its newest version in which a team consists of only two players. They compete with various
other teams. Players were ranked according to their performance in the last 1 year. Four players were selected from India to participate in the tournament. The aggregate
ranks of these players taken two at a time are 49, 36, 64, 57, 43 and 51. If by using one or more of these six mentioned ranks, the coach can uniquely determine the rank of
each player, then what is the difference between the rank of the best and the worst player among them?

To solve the problem, let's denote the ranks of the four players as \(a\), \(b\), \(c\), and \(d\), where \(a \leq b \leq c \leq d\). The aggregate ranks of these players taken two at a time are given as: 49, 36, 64, 57, 43, and 51. The sum of the ranks of each pair is one of these six values.

### Step 1: Sum of All Ranks
The sum of all the pairwise sums can be calculated. This sum would include each player's rank three times (since each player's rank is paired with each of the other three players).

The sum of all pairwise sums:
\[
49 + 36 + 64 + 57 + 43 + 51 = 300
\]
Since each player's rank appears in three sums, the total sum of all the players' ranks can be expressed as:
\[
3(a + b + c + d) = 300
\]
Therefore, the sum of the ranks of the four players is:
\[
a + b + c + d = \frac{300}{3} = 100
\]

### Step 2: Analyzing the Pairwise Sums
Given that \(a \leq b \leq c \leq d\), let's assume:
- \(a + b\) is the smallest sum,
- \(c + d\) is the largest sum.

Identify the smallest and largest sums from the given pairwise sums:
\[
a + b = 36 \quad \text{and} \quad c + d = 64
\]

Now, we have:
\[
a + b = 36
\]
\[
c + d = 64
\]

### Step 3: Finding the Intermediate Sums
Now, using the equation \(a + b + c + d = 100\), substitute the values of \(a + b\) and \(c + d\) into this equation:
\[
36 + 64 = 100
\]
This confirms our choice for \(a + b\) and \(c + d\).

The remaining sums are \(a + c\), \(a + d\), \(b + c\), and \(b + d\). These should match the remaining values \(49\), \(43\), \(57\), and \(51\).

### Step 4: Assigning the Remaining Sums
Given the possible sums, let's try to assign the values:
- Suppose \(a + c = 43\),
- Suppose \(b + d = 57\),
- Suppose \(a + d = 49\),
- Suppose \(b + c = 51\).

### Step 5: Solving for Individual Ranks
Now solve for the individual ranks:

From \(a + b = 36\) and \(a + c = 43\):
\[
b - c = 36 - 43 = -7 \quad \Rightarrow \quad c = b + 7
\]

From \(a + b = 36\) and \(b + d = 57\):
\[
d = 57 - b
\]

From \(c + d = 64\) and \(d = 57 - b\):
\[
c + 57 - b = 64 \quad \Rightarrow \quad c = b + 7
\]
\[
b + 7 + 57 - b = 64
\]

This matches the values calculated for \(c = b + 7\). Now, substitute \(b\) and \(c\) back into the other sums to find specific values:

\[
a + b = 36
\]
\[
b + d = 57
\]
\[
a + d = 49
\]

Finally, let's solve for \(a\), \(b\), \(c\), and \(d\):

\[
b = 15, \quad a = 21, \quad c = 22, \quad d = 42
\]

### Step 6: Determine the Difference
Finally, the difference between the best and the worst player's rank:
\[
d - a = 42 - 15 = 27
\]

Thus, the difference between the rank of the best and worst player is **27**.

---

Would you like more details or have any questions? Here are some related questions:

1. Can you solve a similar problem with different pairwise sums?
2. What happens if the sums provided are not unique, and how do you resolve that ambiguity?
3. How can you verify your solution if the number of players or pairwise sums increases?
4. Is there a quicker method to determine the ranks directly from given sums?
5. How does this approach extend to more than four players?

**Tip:** Always check if the given sum matches exactly when you plug in the values back into the problem; it helps confirm your solution!

Explore how to determine the ranks of four cricket players using given pairwise sums. Understand the application of combinatorics and linear equations to solve the problem effectively. Suitable for high school students and enthusiasts of mathematical problem-solving.

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