Calculating Combinations

[helpingmydaughterwithmath]

My daughter has the following Grade 7 problem to solve:

You have 4 silver coins in your pocket (could be nickels, dimes, quarters)...

How many different combinations of coins could you have?

How many different amounts of money could you have?

She thinks she has the answer to the first question (81). She used 3⁴ to solve it. I'm not much help with either question, as it has been a very long time since I've had to do any serious math .

What's more, her teach left it to the class to "try new things" to solve the questions, without any guidance... I'm pretty sure the second question needs to be solved using factorials/combinations, but damned if I can remember any of the formulas needed. Any assistance would be appreciated!

Thanks!

 

[pka]

My daughter has the following Grade 7 problem to solve:
You have 4 silver coins in your pocket (could be nickels, dimes, quarters)...
How many different combinations of coins could you have?
How many different amounts of money could you have?

Consider this model: \(o,~o,~o,~\&~o\) \(\underbrace {\boxed{|\quad\quad\quad} }_N\underbrace {\boxed{|\quad\quad\quad} }_D\underbrace {\boxed{|\quad\quad\quad} }_Q\)
Now how many ways can we put those four o's into those three boxes?
One way: \(\underbrace {\boxed{|o~o~o~o\quad\quad} }_N\underbrace {\boxed{|\quad\quad\quad} }_D\underbrace {\boxed{|\quad\quad\quad} }_Q\) that is all nickles.
Another: \(\underbrace {\boxed{|\quad o\quad\quad} }_N\underbrace {\boxed{|\quad o\quad\quad} }_D\underbrace {\boxed{|\quad o\quad o\quad} }_Q\) one nickle, one dime & two quarters.
I thinkthat a 7th grader could find all all possible distributions.
I will tell the answer is \(\dfrac{6!}{2!\cdot 4!}=15\) If the class that studied the "stars & bars'" counting principle then the student may understand.

Actually, I would NOT expect a 7th grader to do the second part.
Look at this expansion
The exponents in that expansion tell us the amounts possible the four coins give us.

 

[Dr.Peterson]

My daughter has the following Grade 7 problem to solve:

You have 4 silver coins in your pocket (could be nickels, dimes, quarters)...
How many different combinations of coins could you have?
How many different amounts of money could you have?

She thinks she has the answer to the first question (81). She used 3⁴ to solve it. I'm not much help with either question, as it has been a very long time since I've had to do any serious math .

What's more, her teach left it to the class to "try new things" to solve the questions, without any guidance... I'm pretty sure the second question needs to be solved using factorials/combinations, but damned if I can remember any of the formulas needed. Any assistance would be appreciated!

I didn't know they still made those from silver ... or that there was ever any silver in nickels. But that's okay.

Anyway, with the instruction to "try new things", my guess is that nothing fancy is expected. It might be done by just listing, and watching for patterns to take advantage of. It really shouldn't be done by assuming the answer directly uses combinations or permutations.

The trouble is, the term "combinations" is misleading, and not quite clearly defined here. What your daughter did would be appropriate if order mattered (maybe four slots in a coin holder), so that the first could be one of 3 possibilities, and likewise with the second, third, and fourth. But the coins are in a pocket, which doesn't generally maintain order! So I would interpret the problem as asking for the number of distinct collections possible (not technically sets, because you can have more than one of the same thing).

Since repetition is allowed (indeed, necessary), combinations and permutations are not applicable. So I think you can set aside things you've learned and forgotten, and relax!

I myself would start by starting a list and learning about the problem as I go:

NNNN
NNND
...

Hmmmm ... I don't need to write NNDN, because order doesn't matter. So I can just write the possibilities always in the order NDQ. And that makes this really about finding sums of three numbers that add up to 4, for example NNNN is 4+0+0 and NNND is 3+1+0, and so on.

It turns out that there is a "well-known" formula you could use if you had learned it (and a nice trick that lets you use combinations in an inside-out way, if you forget the formula as I always do) -- but since your daughter surely hasn't, you don't have to worry about that. Just let her know that her answer doesn't really work, and let her "play" with the ideas. I seriously think that a list will be the way to go, for her if not for me. And that will be especially true for the second part.

 

[JeffM]

A hint

How many ways can all the coins be the same? Four: all nickels, all dimes, all quarters, or all half dollars.

How many ways can all the coins but one be the same? Twelve: three nickels, one dime; or three nickels, one quarter; or three nickels, one half dollar; or three dimes, one nickel; or three dimes, one quarter; or three dimes, one half dollar; or three quarters, one nickel; or three quarters, one dime; or three quarters, one half dollar; or three half dollars, one nickel; or three half dollars, one dime; or three half dollars, one quarter.

How many ways can two coins be the same and another two coins be the same? Time for your daughter to think it through.

How many ways can two coins be the same and the other two both different?

How many ways can all four coins be different?

When it comes to value, we have a problem. What are the four types of coins? There are five types if you include both a nickel and a silver dollar.

Without a formula, this is a very tedious question. Even with formulas, it can be tricky because you need to know which formula applies. It is problems like these that turn kids off math.

 

[Deleted member 4993]

It is problems like these that turn kids off math.

I met these problems when I was ~ 12. Actually - these problems made me attracted to math - I could understand the problem statement could not solve it correctly - so near yet so far away. Almost like FLT - almost everybody understood the problem but proving it is a totally different issue.

I am going to give it my grandson. I am sure he is going to get fired up - just to show gram-paw up!!

I am going to be so proud!!

 

[helpingmydaughterwithmath]

Consider this model: \(o,~o,~o,~\&~o\) \(\underbrace {\boxed{|\quad\quad\quad} }_N\underbrace {\boxed{|\quad\quad\quad} }_D\underbrace {\boxed{|\quad\quad\quad} }_Q\)
Now how many ways can we put those four o's into those three boxes?
One way: \(\underbrace {\boxed{|o~o~o~o\quad\quad} }_N\underbrace {\boxed{|\quad\quad\quad} }_D\underbrace {\boxed{|\quad\quad\quad} }_Q\) that is all nickles.
Another: \(\underbrace {\boxed{|\quad o\quad\quad} }_N\underbrace {\boxed{|\quad o\quad\quad} }_D\underbrace {\boxed{|\quad o\quad o\quad} }_Q\) one nickle, one dime & two quarters.
I thinkthat a 7th grader could find all all possible distributions.
I will tell the answer is \(\dfrac{6!}{2!\cdot 4!}=15\) If the class that studied the "stars & bars'" counting principle then the student may understand.

Actually, I would NOT expect a 7th grader to do the second part.
Look at this expansion
The exponents in that expansion tell us the amounts possible the four coins give us.

Thanks PKA, but can you explain how you got the values for your combination formula? If C = n!/r!(n-r)!, wouldn't n = 4?

And I'm completely lost with the second part of the problem (solving for the possible sums)... Like I said, I've been out of the classroom for a long time Also, my mistake, I hit the wrong key when I typed my original post - my daughter is actually in Grade 8

 

[helpingmydaughterwithmath]

A hint

How many ways can all the coins be the same? Four: all nickels, all dimes, all quarters, or all half dollars.

How many ways can all the coins but one be the same? Twelve: three nickels, one dime; or three nickels, one quarter; or three nickels, one half dollar; or three dimes, one nickel; or three dimes, one quarter; or three dimes, one half dollar; or three quarters, one nickel; or three quarters, one dime; or three quarters, one half dollar; or three half dollars, one nickel; or three half dollars, one dime; or three half dollars, one quarter.

How many ways can two coins be the same and another two coins be the same? Time for your daughter to think it through.

How many ways can two coins be the same and the other two both different?

How many ways can all four coins be different?

When it comes to value, we have a problem. What are the four types of coins? There are five types if you include both a nickel and a silver dollar.

Without a formula, this is a very tedious question. Even with formulas, it can be tricky because you need to know which formula applies. It is problems like these that turn kids off math.

Thanks Jeff, but the only coins in the pocket are any combination of nickels, dimes and quarters (no half-dollars). When my daughter realized her original answer was wrong, she started working it out the long way, like both you and Dr. Peterson noted... NNNN, NNND,...

However, we are both still completely lost on the second half of problem (how many possible sums are there?).

Oh, and I forgot to mention, the teacher gave her a second problem... Same two questions, but this time 10 coins in the pocket! I don't think she's going to be able to figure that one out (at least the sums question) without a formula of some sort

 

[Dr.Peterson]

Thanks PKA, but can you explain how you got the values for your combination formula? If C = n!/r!(n-r)!, wouldn't n = 4?

And I'm completely lost with the second part of the problem (solving for the possible sums)... Like I said, I've been out of the classroom for a long time Also, my mistake, I hit the wrong key when I typed my original post - my daughter is actually in Grade 8

What pka showed you is what I called "a well-known formula" (not a straightforward application of combinations, but indirect), and "stars-and-bars is my "nice trick" by which you can derive it -- both presumably far beyond what your daughter has learned. At least, I'm guessing she has learned the product rule, by which she decided on 3^4, and maybe (or maybe not) permutations or combinations, but nothing beyond that. If she were writing to us, I would have asked her what she has learned, and try specifically to help her use that, if applicable. As it is, I can't be sure how to help her (or you) most effectively, or what is expected.

But the second part, I think, has to be done by listing possibilities; there can't be a formula for that, other than what pka showed you, which is definitely too advanced and still doesn't give a count. And since, as pka has told you, the list will have only 15 items on it, I say a list is probably what is expected.

An exploration like this, without a simple formulaic solution, is valuable for students in showing that the formulas they are taught are not enough to solve every problem; deep thinking is needed, which is what makes this field fun! Without tools, it can be discouraging, especially if it isn't made clear that something other than a formula is appropriate. But when you realize that you can find an answer on your own, even if it doesn't feel very "mathematical", it can be exhilarating.

 

[pka]

We are asked to find the number of ways that four coins can be selected from nickles. dimes & quarters.
Asked to find the number of ways that four balls from red, blue & green.
Asked to find the number of ways that four people from Greeks, Arabs & English.
All three have the same answer: the number of ways to arrange the string \(o,~o,~o,~o,~| ,~|~\)
That string can be rearranged in \(\dbinom{6}{2}=\dfrac{6!}{2!\cdot 4!}=15\)
In general: asked to find the number of ways that \({\bf N}\) selections from \({\bf K}\) types
That answer is \(\dbinom{N+K-1}{N}\)

 

[Steven G]

Whenever a student gives me to check their answer to a counting problem I ask them if they can list the outcomes (as long as it is not too long). The reason is that student do not have to ask their professor if they counted correctly as they could make a list and see for themselves if they are correct or not. When they are wrong then can see where they went wrong! Listing 15 combinations is not too hard. Part b is actually easy if you have the list! Just add up the sum for each entry on the list and count the number of different values. I thing that only 40 cents can come up more than once--as qnnn and dddd.

 

[pka]

Whenever a student gives me to check their answer to a counting problem I ask them if they can list the outcomes (as long as it is not too long). The reason is that student do not have to ask their professor if they counted correctly as they could make a list and see for themselves if they are correct or not. When they are wrong then can see where they went wrong! Listing 15 combinations is not too hard. Part b is actually easy if you have the list! Just add up the sum for each entry on the list and count the number of different values. I thing that only 40 cents can come up more than once--as qnnn and dddd.

To see what Jomo is about see this:
\(\begin{array}{*{20}{c}}
N&\ & D&\ & Q \\


\hline


4&\ & 0&\ & 0 \\


3&\ & 1&\ & 0 \\


3&\ & 0&\ & 1 \\


2&\ & 2&\ & 0 \\


2&\ & 0&\ & 2 \\


2&\ & 1&\ & 1 \\


1&\ & 0&\ & 3 \\


1&\ & 1&\ & 2 \\


1&\ & 2&\ & 1 \\
1&\ & 3&\ & 0 \\
1&\ & 0&\ & 3 \\
0&\ & 0&\ & 4 \\
0&\ & 4&\ & 0 \\
0&\ & 1&\ & 3 \\
0&\ & 2&\ & 2 \\
0&\ & 3&\ & 1
\end{array}\)