Help me do my second grader's math homework

Factual Questions
1

Last night my daughter brought home a math assignment that sort of stumped us. The problem was to take 5 flavors of ice cream (say, vanilla, chocolate, strawberry, orange, and lime) and combine them into triple dip cones.

Now, if the problem had been to find all the possible permutations, I have some idea how to diagram it systematically. However the rule was that, while you could have more than one scoop of the same flavor in a cone, or even all three the same, you were only supposed to have one version of each combination. That is, vanilla/chocolate/strawberry was to be considered the same as chocolate/vanilla/strawberry, etc.

We started by listing all the flavors three of the same at a time, then the combinations with two scoops the same, then the rest. I think we got them all, but it felt a bit random and we weren’t sure exactly what she was supposed to be getting out of all this.

Any ideas about a better approach that a 7-year-old could grasp?

2

There is a formula for doing this easily, but obviously, simple as it is, it’s going to be over the head of a 7 year old. I’d imagine they just wanted her to see that Choc-Van-Straw is the same as Van-Straw-Choc is the same as Straw-Van-Choc, etc. Although it does seem like a pain to get that point across.

3

If you are explaining it to a 7 year old, you might say “if you had to choose 3 out of 5 people to be on a team or in a club or whatever, how many ways could this be done?” The answer being 10 and it could be shown easily.

However, when you say that the same person (or ice cream flavor) can be used more than once, I do not know what principle they would be attempting to show a 7 year old.

4

Well, the number of combinations is [sup]5[/sup]C[sup]2[/sup] or

(5)
(3)

(where each two vertical brackets represent one larger vertical bracket), both symbols being defined to mean “the number of ways of choosing three things from five”.

The number is 5x4x3/3!=5x4x3/3x2x1=10. http://en.wikipedia.org/wiki/Combinations_and_permutations provides a short introduction.

As for actually listing them, the traditional approach is to list all the ones with chocolate in, (fairly easy, as you’re choosing two from four, or do the same trick with two of the remaining flavours), and then all the ones without chocolate but with vanilla, etc.

5

My suggestion is to go to the Quik Trip, lay out the big bucks for the ice cream and start scooping.

Your daughter may not learn any math, but it will be a lot fun. Unless she actually tries to consume all the ice cream, in which case you will save a lot of money in the future, as she won’t ask for ice cream ever again.

6

If order makes a difference, that the number of possible combinations are 125 (i.e., 5 x 5 x 5). Clearly with order ignored, you have fewer combinations.

There are three possibilities:
(a) All three flavours the same. There are 5 ways of doing this.
(b) One flavour repeated, with one other flavour. There are 20 ways of doing this (5 choices for the repeated flavour times 4 choices for the other flavour)
© Three different flavours. If order makes a difference, then there are 5 x 4 x 3 different combinations, and then you have to allow for order not making a difference. However, a way to make the calculations easier in this case is to look at what flavours are being left out. Of the 5 flaours, if you choose 3, then 2 are left out. So you have 5 x 4 combinations (if order matters). But order does not matter, so chocolate-vanilla is the same at vanilla-chocolate. Therefore, you divide 5 x 4 by 2, giving 10 combinations.

The answer: 5 + 20 + 10 = 35

7

These aren’t permutations, but combinations.

Given a number of elements in a set (call it “n”), and taking combinations of “k” items, the term is [sub]n[/sub]C[sub]k[/sub].

The formula is:



 (n+k-1)!
k!*(n-1)!

So, in your example, n=5 and k=3, so you get 7!/(3!4!), or 5040/(624), or 5040/144 or 35. I think.

But in this case, the intent was probably to do it by trial and error just to grasp the concept, as audiobottle said.

8

Thanks, everyone. The way Giles described it is pretty much the method we hit on. I appreciate having the formula, though. It even looks familiar. I can see I need to get back up to speed on this stuff before she hits the big time. I thought I’d have a few more years to review, though!

9

That’s not really an appropriate problem for a second-grader, IMO. The formula is given here, but I have no idea how to explain it to a seven-year old kid.

FYI, there are 35 possibilities. Listing all of them without knowing how many to look for would be tough.

10

The trick to doing this, without using the formula, is to cycle through systematically. (As an aside, what ultrafilter said. I’ve got a PhD, in engineering no less, and I have to think slowly and carefully in order to make sure I’m doing this right. I can’t imagine how this is appropriate for someone who is seven years old.)

Anyway, list the flavors in some order; let’s say chocolate, vanilla, strawberry, orange, and lime – or C, V, S, O, L.

Start with all chocolate (CCC). Combination one. Then cycle through the last position: CCV, CCS, CCO, and CCL. Five total now, and these are all the combinations with two chocolate scoops.

Now (and here’s the trick), find all the combinations which have chocolate and vanilla. There are only four additional, because you already counted CCV. The additional ones are CVV, CVS, CVO, and CVL. Note I’m cycling through the last position using the same order.

Then find the three combinations with C and S, the two with C and O, and the one with C and L.

And those are all the combinations with chocolate. Throw the chocolate away and start in on the rest of the possibilities. When you’re done, you’ll have a list like this:



CCC
CCV
CCS
CCO
CCL

CVV  VVV
CVS  VVS
CVO  VVO
CVL  VVL

CSS  VSS  SSS
CSO  VSO  SSO
CSL  VSL  SSL

COO  VOO  SOO  OOO
COL  VOL  SOL  OOL

CLL  VLL  SLL  OLL  LLL

Note the repetitions of the pattern.

11

PS. Ignore my calculation, I didn’t see that two scoops could be the same. :smack:

Is it appropriate? If you want them to take an hour and miss a couple, maybe. If not, maybe not. I don’t know if their homework would expect that, but I remember that when I was at school, when I was younger homework was more “Have a play with this” and older “You WILL do this.” I have no idea what age ‘grade 7’ is :slight_smile: For that matter maybe she’s supposed to ask mummy to remember the total number, that’d make it easier.

12

Or you could draw it as a decision ladder, like so:


A B C
|_|_|
| | |
A B C
|_|_|
| | |
A B C

Then let your daughter trace the possible paths through the ladder. However, since the order of the letter’s don’t matter, you can save yourself a few trips.

13

We weren’t exactly sure what the idea is. Her shiny new teacher gives weekly reports on what they are doing, though, so maybe he will shed some light on it. It’s definitely nothing to do with formulas, though, because they haven’t even reached multiplication and division in any formal way yet.

My guess is the purpose was: 1) To get an idea that there are lots more combinations. They did a similar problem with two scoops previously. 2) To introduce the idea of solving problems systematically, rather than just plunging in with random combinations, and 3) To make sure that Mommy and Daddy get involved in your homework once in a while beyond just checking your arithmetic and making sure you don’t make your 3’s backwards.

14

Maybe you could explain it like this. To make the problem easier to compute, I’ll assume that you have only 3 flavors. But this technique easily generalizes to any number of flavors.

Make a chart with columns corresponding to flavors and rows corresponding to ways to make the cone. You will have three columns, labeled C, S, and V, respectively. You don’t know yet how many rows you will need, since that is what you are trying to find. So, you start out with something like this:



-------------
| C | S | V |
-------------
-------------
|   |   |   |
-------------
|   |   |   |
-------------

Now, in each row, write in each column the number of scoops you have of the corresponding flavor. The only condition you need to satisfy is that the sum of the entries in each row should be 3.

You can list these possibilities systematically like this:



-------------
| C | S | V |
-------------
-------------
| 0 | 0 | 3 |
-------------
| 0 | 1 | 2 |
-------------
| 0 | 2 | 1 |
-------------
| 0 | 3 | 0 |
-------------
| 1 | 0 | 2 |
-------------
| 1 | 1 | 1 |
-------------
| 1 | 2 | 0 |
-------------
| 2 | 0 | 1 |
-------------
| 2 | 1 | 0 |
-------------
| 3 | 0 | 0 |
-------------

Now you count the number of rows to get the answer: 10.

15

That method could be useful Tyrell McAllister. Based on some other recent assignments they seem to be trying to get the kids to think about how to present information graphically. This was a preprinted workbook exercise and the original instructions involved using graph paper and colored blocks to work it out.

16

OK, I seem to be having a dense moment here.

I can see logically that the answer is 35, and the equation for a combination gives 35, but can someone explain why my calculator gives “10” as the answer for 5C3???

17

Because the number 5C3 is the number of ways to make the cone without repetition. Since you are allowed to repeat flavors, there are more ways to make the cone.

18

Here’s a thread from last month where I tried to answer the same question, using the method that was explained to me when I was a sophomore in college.

19

Yep, thanks. I figured that out after looking over the Wiki page formulas. I guess something to keep in mind with my calculator at least.

At any rate, this seems WAY advanced for a 2nd grader.

20

Is the problem to list all combinations (used loosely here) or to count them? I don’t see how one way of listing them by hand is more efficient than another. Creating a computer program to do it is another matter; there may be some algorithms that are more efficient than others. I haven’t really thought about it. However, since humans can simply eyeball whether two combinations are the same, I don’t see how one way is more efficient than another.

To see why the formula that others have listed works, consider the following. You want to find the number of solutions to the equation

x_1 + x_2 + x_3 + x_4 + x_5 = 3

where x_1, x_2, x_3, x_4, x_5 are natural numbers (including zero).

This is the same as the number of solutions to

y_1 + y_2 + y_3 + y_4 + y_5 = 3 + 5

where y_1, y_2, y_3, y_4, y_5 are positive integers.

The number of solutions to the latter equation can be counted as follows. Suppose you line up 8 balls in a row. There are 7 spaces in between the balls. Let ‘o’ represent a ball and ‘|’ represent a space. Then, the situation looks like this.

o|o|o|o|o|o|o|o

You want the number of ways to divide these balls into 5 segments, where each segments contains at least one ball. Dividing them into 5 segments is the same as choosing 4 of the spaces. The number of ways to do this is

C(7,4) = 7!/(4!*3!) = C(5+3-1,3)

This can be easily generalized to n flavors and k scoops.

I don’t see how you can explain this to a 2nd-grader who doesn’t know multiplication and division.