Mathematical Insights: How Many Ways Can Coins Be Distributed to Boys?

Imagine a scenario where a man has a specific set of coins and wants to distribute them to a group of boys such that each boy receives one coin. This intriguing problem not only tests our understanding of combinatorics but also provides a practical application of mathematical principles. Let's dive into the details to discover how many ways the distribution can be achieved.

Step 1: Understanding the Problem Statement

The man has:

1 five-peso coin 2 one-peso coins 3 twenty-five-centavo coins 4 ten-centavo coins

With a total of 10 coins, he needs to distribute these to 10 boys, ensuring each boy gets exactly one coin. This means that the coins are distinct in terms of their value, even though some are indistinguishable among themselves. The challenge is to calculate the number of different ways to distribute these coins among the boys.

Step 2: Applying the Multinomial Coefficient

The problem of distributing these coins can be treated as a combinatorial problem. We will use the multinomial coefficient to determine the number of ways to distribute the coins. The formula for the multinomial coefficient is given by:

(frac{n!}{n_1! cdot n_2! cdot n_3! cdots n_k!})

where ( n ) is the total number of items to distribute, and ( n_1, n_2, n_3, ldots, n_k ) are the counts of each distinct item.

Step 3: Calculating the Values

In this specific scenario:

( n 10 ) (total coins) ( n_1 1 ) (five-peso coin) ( n_2 2 ) (one-peso coins) ( n_3 3 ) (twenty-five-centavo coins) ( n_4 4 ) (ten-centavo coins)

We can plug these values into the formula:

(text{Number of ways} frac{10!}{1! cdot 2! cdot 3! cdot 4!})

Step 4: Calculating the Factorials

Let's compute the factorials first:

10! 3,628,800 1! 1 2! 2 3! 6 4! 24

Substituting these factorials back into the formula, we get:

(frac{3,628,800}{1 cdot 2 cdot 6 cdot 24} frac{3,628,800}{288})

Step 5: Final Calculation

Performing the final calculation:

(frac{3,628,800}{288} 12,600)

Conclusion

Hence, the number of ways the boys can profit by each receiving one coin is 12,600. This result encapsulates the unique distribution possibilities of the given set of coins among the boys, demonstrating the power of combinatorial mathematics in solving practical problems.

Note: It's important to recognize that the calculation accounts for the different types of coins and their counts. The probability aspect of the problem is not directly addressed here, as the focus is on the number of distinct distributions.