There are 210 boys attending the national cricket academy at Bangalore. Each of them is assigned to work with one of the twenty coaches. It is noted that each coach works with a distinct distinct number of boys. In how many ways can the groups of boys be assigned?

Answer

To solve the problem of assigning 210 boys to 20 coaches such that each coach works with a distinct number of boys, we can follow these steps:

### Step 1: Understanding the Distribution
We need to assign 210 boys to 20 coaches in such a way that each coach has a different number of boys. This means we need to find a combination of distinct positive integers that sum up to 210.

### Step 2: Finding Distinct Positive Integers
The smallest 20 distinct positive integers are 1, 2, 3, ..., 20. The sum of these integers can be calculated using the formula for the sum of the first n natural numbers:

\[ S = \frac{n(n + 1)}{2} \]

For \( n = 20 \):

\[ S = \frac{20 \times 21}{2} = 210 \]

This means we can assign the boys such that Coach 1 gets 1 boy, Coach 2 gets 2 boys, ..., and Coach 20 gets 20 boys.

### Step 3: Assigning the Boys
Now that we have the distribution of boys to coaches, we need to calculate the number of ways to assign the boys. The number of ways to assign the boys can be calculated using the multinomial coefficient:

\[ \frac{n!}{k_1! \cdot k_2! \cdot k_3! \cdots k_{20}!} \]

Where \( n \) is the total number of boys (210), and \( k_i \) is the number of boys assigned to each coach.

In our case, we have:

- Coach 1: 1 boy
- Coach 2: 2 boys
- Coach 3: 3 boys
- ...
- Coach 20: 20 boys

Thus, we can express this as:

\[ \frac{210!}{1! \cdot 2! \cdot 3! \cdots 20!} \]

### Step 4: Final Calculation
Now we can compute the final answer:

\[ \text{Number of ways} = \frac{210!}{1! \cdot 2! \cdot 3! \cdots 20!} \]

This gives us the total number of ways to assign the boys to the coaches under the given constraints.

### Summary
The number of ways to assign 210 boys to 20 coaches, where each coach has a distinct number of boys, is given by:

\[ \frac{210!}{1! \cdot 2! \cdot 3! \cdots 20!} \]