Answered

In each of the parts below, the coins available are half-dollars, quarters, dimes, nickels, and pennies.


AConsider the ways in which you can make a dollar with 6 coins.
  • How many different ways can you make a dollar using exactly 1 half-dollar?
  • How many different ways can you make a dollar using 0 half-dollars?
  • How many different ways can you make a dollar with 6 coins?


BConsider the ways in which you can total 95 cents with 7 coins.
  • How many different ways can you total 95 cents using exactly 1 half-dollar?
  • How many different ways can you total 95 cents using 0 half-dollars?
  • How many different ways can you find for 7 coins totaling 95 cents?


CConsider the ways in which you can total one dollar with 21 coins.
  • How many different ways can you total one dollar using exactly 1 half-dollar?
  • How many different ways can you total one dollar using exactly 2 quarters?
  • How many different ways can you total one dollar using exactly 1 quarter?
  • How many different ways can you total one dollar using 0 quarters and 0 half-dollars?
  • How many different ways can you find for 21 coins totaling one dollar?

Verified Answer :

  • Part (a) 100¢ with 6 coins: 1H (\(2\)), 0H (\(1\)), Total (\(3\)).
  • Part (b) 95¢ with 7 coins: 1H (\(1\)), 0H (\(2\)), Total (\(3\)).
  • Part (c) 100¢ with 21 coins: 1H (\(1\)), 2Q (\(1\)), 1Q (\(1\)), 0Q&0H (\(2\)), Total (\(5\)).
  • Final Answers: (a) \(\boxed{2, 1, 3}\); (b) \(\boxed{1, 2, 3}\); (c) \(\boxed{1, 1, 1, 2, 5}\).

Explanation

  1. Define the mathematical model and variables

We represent the number of half-dollars, quarters, dimes, nickels, and pennies as \(h\), \(q\), \(d\), \(n\), and \(p\), respectively. Each coin has a specific value: \(H=50\), \(Q=25\), \(D=10\), \(N=5\), and \(P=1\). The constraints provided are the total value in cents and the total number of coins. We use the two fundamental equations:

  1. Total Value: \(50h + 25q + 10d + 5n + p = \text{Total}\)
  2. Total Coins: \(h + q + d + n + p = N_{\text{coins}}\)

By substituting specific constraints (e.g., \(h=1\)), we can reduce these equations to solve for the remaining variables.

  1. Analyze Part (a): 100 cents with 6 coins

For the case of making \(100\) cents with \(6\) coins:

  • For \(1\) half-dollar (\(h=1\)): The equations become \(q+d+n+p=5\) and \(25q+10d+5n+p=50\). Subtracting the equations gives \(24q+9d+4n=45\). The valid integer solutions for \((q,d,n)\) are \((0,5,0)\) (giving \(p=0\)) and \((1,1,2)\) (giving \(p=1\)). Thus, there are \(2\) ways.
  • For \(0\) half-dollars (\(h=0\)): The equations are \(q+d+n+p=6\) and \(25q+10d+5n+p=100\). Subtracting yields \(24q+9d+4n=94\). The only valid solution is \((3,2,1)\) (giving \(p=0\)). Thus, there is \(1\) way.
  • The total number of ways is \(2 + 1 = 3\).
  1. Analyze Part (b): 95 cents with 7 coins

For the case of making \(95\) cents with \(7\) coins:

  • For \(1\) half-dollar (\(h=1\)): The equations are \(q+d+n+p=6\) and \(25q+10d+5n+p=45\). Subtracting gives \(24q+9d+4n=39\). The only valid solution is \((1,1,1)\) (giving \(p=3\)). Thus, there is \(1\) way.
  • For \(0\) half-dollars (\(h=0\)): The equations are \(q+d+n+p=7\) and \(25q+10d+5n+p=95\). Subtracting gives \(24q+9d+4n=88\). Valid solutions are \((3,1,1)\) (giving \(p=2\)) and \((2,4,0)\) (giving \(p=1\)). Thus, there are \(2\) ways.
  • The total number of ways is \(1 + 2 = 3\).
  1. Analyze Part (c): 100 cents with 21 coins

For the case of making \(100\) cents with \(21\) coins:

  • For \(1\) half-dollar (\(h=1\)): There is \(1\) way.
  • For \(2\) quarters (\(q=2\)): There is \(1\) way.
  • For \(1\) quarter (\(q=1\)): There is \(1\) way.
  • For \(0\) quarters and \(0\) half-dollars (\(q=0, h=0\)): There are \(2\) ways.
  • The total number of ways to make \(100\) cents with \(21\) coins is \(5\).

Examples

This algebraic approach to counting combinations of coins is a fundamental exercise in combinatorial optimization. It is used in real-world scenarios such as designing vending machine change-making algorithms, where the goal is to determine the most efficient (or all possible) ways to return change using the fewest coins possible or specific denominations. Mastering this method allows for quick calculations in everyday financial contexts and forms the basis for more complex partition problems in computer science and discrete mathematics.

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