Combinatorial Approach to Building a Dollar

The entire modern world relies on combinatorics. Combinatorics is the branch of mathematics studying finite or countable discrete structures (ESSENTIALLY, HOW MANY WAYS CAN YOU ORDER A SET OF OBJECTS USING A PARTICULAR SET OF RULES?).

Without this branch of mathematics, computer programs that take a slit second would take weeks. Cellphone communications (error correcting codes, wavelet and fourier optimizations), game programing (polygon optimization), probability and statistics, and every website that has over 100 pieces of data, requires combinatorics. Now, how do we make this understandable to our children?

How many ways (combinations) can we create a dollar with pennies, nickels, dimes, quarters, half dollars, and dollar coin? The only way to accomplish this task efficiently is to find a pattern or system of recording the solutions.

The youngest Mathletes should just focus on using this algebraic notation below to make any combination of one dollar even if it is random. Please have them use actual coins when doing this exercise to give them a push but they should record their solutions. Do not worry if their solutions do not equal one dollar; this gives us an opportunity to understand why they missed it and how to help them develop better number sense.

WOULD YOU BELIEVE 293 DIFFERENT WAYS

How do we represent this question algebraically? With Algebraic Notation of course (Substitute the number of coins used for the appropriate expression):

P = pennies

N = nickels

D = dimes

Q = quarters

H = half-dollar coin

W = whole dollar coin

1W = 100

2H = 50 + 50

1H 2Q = 50 + 25 + 25

1H 1Q 2D 1N = 50 + 25 + 10 + 10 + 5

1H 1Q 2D 5P = 50 + 25 + 10 + 10 + 5

1H 1Q 1D 3N = 50 + 25 + 10 + 5 + 5 + 5

1H 1Q 1D 2N 5P = 50 + 25 + 10 + 5 + 5 + 5

I started with an easy challenge. How many ways (combinations) can you make 100 with just quarters and dimes?

0Q 10 D

2Q 5D

4Q 0D

I LIKE USING ZEROS TO KEEP THINGS EASY STARTING WITH THE ZERO OF THE LARGEST COIN VALUE

ONLY 3 WAYS???? WOW, THAT WAS EASY

Now, let’s add a level of complexity (make it harder). How many ways (combinations) can you make 100 with just quarters, dimes, and nickels? 

First, I asked each child to give me one solution and wrote it down on the white board. I asked them if these solutions were in an order. They saw the random nature of their solutoins. Then, turning over the board, I asked them to come up with a systematic order so we find all solutions without fail.

There are 29 ways and your job is to find them all but you have to do this systematically (in order) so you don’t get confused.

0Q 0D 20N

0Q       1D 18N

0Q       2D 16N

Then I asked them what would happen when we got to:

0Q        10D  0N

Many of them said 4Q + 0D + 0N but then they said, "no, let's start with 1Q like we followed 0D up above with 1D then 2D, etc.

1Q        0D  15N

1Q        1D  13N

1Q        2D  11N

Then, follow this sequence until you reach the final solution of:

4Q        0D  0N

CRAZY CHALLENGE (FOR GRADES 3-49 [yes, I am in 49th grade]): 

Now, let’s add another level of complexity (make it even harder). How many ways (combinations) can you make 100 with just quarters, dimes, and nickels and pennies? 

There are 242 ways and your job is to find them [all] but you have to do this systematically (in order) so you don’t get confused. Do not get frustrated trying to find them all. Just have fun trying to come up with a system. On the attached pdf, there are exactly 242 rows for recording solutions and the last row starts with 4Q.

0Q 0D 0N 100P

0Q  0D  1N 95P

0Q  0D  2N 90P

The big challenge here is what to do after you reach:

0Q  0D  20N 0P

And, we now need to add 1D so:

0Q  1D  0N 90P

0Q  1D  1N 85P

And, as you look at these solutions, you can see that the position of the first line with 1Q will be the 112th. Some of the older Mathletes solved this by adding the first grouping of pennies of 20 + 18 + 16 + .... + 0 = 110 but they saw the 0P line as an additional line 111 so the first line with 1Q would be the 112th. Now the challenge is to find the position with 2Q, then 3Q, and of course the 242nd line with 4Q.