The Combinatorial Challenge of Forming Different Sums with Unique Coins

Imagine a girl with five coins of different denominations. The question arises: how many different sums of money can she form? This problem involves a blend of combinatorial mathematics and practical application, providing a fascinating exercise in logical thinking and problem-solving.

Distinct Denominations and Possible Combinations

Each coin in this scenario has a unique value. Let's denote the denominations as a1, a2, a3, a4, a5. To solve the problem, we first recognize that for each coin, the girl has two choices: either to include it in the sum or exclude it. This combinatorial approach simplifies the process of determining all possible sums.

Total Combinations and Excluding the Empty Set

The total number of combinations of the five coins can be calculated using the formula (2^n), where (n) is the number of coins. In this case, (n 5), so the total number of combinations, including the empty set, is:

[2^5 32]

Since the problem specifies that we need to exclude the empty set (a sum of 0), we subtract 1 from the total number of combinations:

[32 - 1 31]

Therefore, the girl can form 31 different sums of money using her five unique coins of different denominations.

Manual Verification with Letters and Values

To verify the solution, let's use letters A, B, C, D, and E representing the coins. We can list all possible combinations and see how many unique sums we can form:

No coins: 1 combination (empty set) One coin: 5 combinations (A, B, C, D, E) Two coins: 10 combinations (AB, AC, AD, AE, BC, BD, BE, CD, CE, DE) Three coins: 10 combinations (ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE) Four coins: 5 combinations (ABCD, ABCE, ABDE, ACDE, BCDE) Five coins: 1 combination (ABCDE)

Adding these combinations together, we get:

[1 5 10 10 5 1 32]

Since we exclude the empty set, the total number of unique sums is 31.

Applying Specific Values

Lets verify the solution using actual coin values, with the coins being 1 cent, 5 cents, 10 cents, 25 cents, and 50 cents. The possible sums are calculated as follows:

0.00 (no coins) 0.01 (A), 0.05 (B), 0.10 (C), 0.25 (D), 0.50 (E) 0.06 (AB), 0.11 (AC), 0.15 (AD), 0.26 (BC), 0.30 (BD), 0.31 (BE), 0.35 (CD), 0.40 (CE), 0.41 (DE), 0.56 (ABE) 0.16 (ACB), 0.36 (BCE), 0.66 (ABDE), 0.60 (ACDE), 0.65 (BCDE) 0.31 (ABCDE)

Counting these distinct sums, we find a total of 31 unique values.

Conclusion

The exercise of determining the number of unique sums that can be formed with five coins of different denominations is a classic problem in combinatorics. By understanding the concept of combinations and including all possible sums, we can solve the puzzle effectively. The problem provides a practical application of combinatorial mathematics and highlights the importance of considering every possible scenario, even when seemingly straightforward at first glance.