How many different ways can you make change for a quarter? (Different arrangements of the same coins are not counted separately.)

Maiclubk

Maiclubk

Answered question

2020-11-29

How many different ways can you make change for a quarter?
(Different arrangements of the same coins are not counted separately.)

Answer & Explanation

Liyana Mansell

Liyana Mansell

Skilled2020-11-30Added 97 answers

There are three coins that are less than a quarter (0.25$) in value,that is a penny (0.01$), a nickel (0.05$) and a dime (0.10$). 
Possibilities are then: 
2 dimes 1 nickel, 2 dimes and 5 pennies, 1 dime and 3 nickels, 1 dime and 2 nickels and 5 pennies, 1 dime and 1 nickel and 10 pennies, 1 dime and 15 pennies, 5 nickels, 4 nickels and 5 pennies, 3 nickels and 10 pennies, 2 nickels and 15 pennies, 1 nickel and 20 pennies, 25 pennies 
Answer is 12

alenahelenash

alenahelenash

Expert2023-06-14Added 556 answers

Answer: 9
Explanation:
1. If we use a quarter as the first coin, the remaining amount to be made is (25 - 25) = 0. In this case, there is only one way to make change: using a single quarter.
2. If we use a dime as the first coin, the remaining amount to be made is (25 - 10) = 15. We can determine the number of ways to make change for 15, denoted as C(15), using the same method.
3. If we use a nickel as the first coin, the remaining amount to be made is (25 - 5) = 20. We can determine the number of ways to make change for 20, denoted as C(20), using the same method.
4. If we use a penny as the first coin, the remaining amount to be made is (25 - 1) = 24. We can determine the number of ways to make change for 24, denoted as C(24), using the same method.
To find C(), we sum up the number of ways for each case:
C()=C(0)+C(15)+C(20)+C(24)
Using the same approach, we can break down each subproblem until we reach base cases that can be directly computed. The base cases are:
C(0)=1 (only one way to make no change)
C(5)=1 (only one way to make change with a nickel)
C(10)=1 (only one way to make change with a dime)
C(15)=2 (two ways: 1 dime and 1 nickel, or 3 nickels)
C(20)=2 (two ways: 2 dimes, or 1 dime and 2 nickels)
C(24)=4 (four ways: 2 dimes and 4 pennies, or 1 dime, 2 nickels, and 4 pennies, or 3 nickels and 9 pennies, or 1 dime and 14 pennies)
Now, we can substitute these values back into the original equation:
C()=1+2+2+4=9
Therefore, there are 9 different ways to make change for a quarter.
karton

karton

Expert2023-06-14Added 613 answers

We need to find the number of solutions for the equation 25q+10d+5n+p=25, where q,d,n,p are non-negative integers.
Using generating functions, we can express this equation as:
(1+x+x2+)(1+x10+x20+)(1+x5+x10+)(1+x)
The coefficient of x25 in this expression represents the number of ways to make change for a quarter.
Simplifying the expression, we have:
1(1x)(1x10)(1x5)(1x)
Expanding this expression using partial fractions, we can write:
A1x+B1x10+C1x5+D1x
Solving for the values of A,B,C, and D, we can substitute them back into the expression and find the coefficient of x25.
Thus, the number of different ways to make change for a quarter is the coefficient of x25 in the expanded expression.

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