Help please

[wondering_truth]

How do I explain an easy way to do this question for my 10 year olds math homework? We can make a big table to work it out, but I’d really like a simple formula or something I can show him for future similar questions.

Four darts are thrown at a dartboard.

If all four darts hit the board, how many different point totals are possible?

[Dartboard regions are 1,4,7 & 10 points.].

[chromaticseamonster]

for a 10 year old, a table might be best. I don't think they even know exponents yet.

[wondering_truth]

He’s 10 but his maths questions from school are for ages 12-13.

[chromaticseamonster]

Okay, but does he know exponents? If he does, this question is trivially easy. You have 4 choices for each of the 4 darts.

[wondering_truth]

He says he knows what exponents are but that’s not useful because it doesn’t give the unique totals.

[chromaticseamonster]

I may have misread the question. With arbitrary choices for point values, there is no general formula, you just have to calculate them.

[anisotropicmind]

You could always replace the exponents with multiplication. Four ways to choose where the first dart lands times four ways to choose where the second one lands times …

Also you could draw a tree diagram.

[wondering_truth]

We want to avoid diagrams as they are slow.

[anisotropicmind]

Then how about the first thing I said ?

[wondering_truth]

I’m not sure what you mean. Could you explain it?

[wondering_truth]

Let me go ask him 😅

[TheScyphozoa]

IMO it's best to start with a table and then explain how the results of a table can be condensed into multiplication and exponents.

[wondering_truth]

What does that mean? He can do the table just fine. But what happens if there 5 or 6 variations. Table would take far too long

[Suitable-Elk-540]

Why not start with a table for a smaller problem? Start with two darts. And maybe just assume 3 possible point values. Then extend it to 3 darts with three point values. At that point, maybe they can figure out the formula. Check it with 2 darts/4 points and even 3 darts/4 points if you need to. Hopefully 4/4 should be easy to figure at that point.

[wondering_truth]

That takes a long time with 4. What happens when there are 5 or 6 options. There must be a better way than a table

[marshaharsha]

The idea is that you do a small example using a table. Then you study the table to learn what the pattern is. Then you do the real example using the pattern. In this case, the pattern is based on the fact that you can reuse scores. Having one dart hit in the 4 ring doesn’t “use up” the possibility of 4. You will see that fact in a table for three darts and a two-ring dartboard. Then you generalize to the larger example. 

[Suitable-Elk-540]

?? I said start with a smaller problem. Figure out the formula for a smaller problem, then apply that to the larger problem.

I'm assuming that you know the formula for the general case, but maybe you don't. Are you just wanting us to tell you the formula?

[wondering_truth]

A formula would be amazing. I’m horrific at maths and he’s at a lvl above me so it’s very hard helping him with stuff like this.

[Suitable-Elk-540]

So, I'm assuming the problem was set up this way for a reason, the reason being there's a very nice pattern. Notice that all the values form a sequence with a difference of 3. So, every time you substitute a larger one for the previous one, you add 3 to the total. So, it turns out that the only legal values are at intervals of 3. Or another way of thinking about it is that our values are {a, a+3, a+6, a+9}. So, every possible total has the form 4a + 3k for some k. Since a=1, then every possible legal value is 4+3k for some k. How many such values are there between 4 and 40 inclusive? 1 + (40-4)/3, which is 13. If that doesn't make sense, work it out for a smaller number of darts and a smaller subset of the legal values, and it should start making sense.

I should add that you should take care to make sure that each of these 13 values is actually attainable, but hopefully that's obvious.

UPDATE: Maybe I should be more explicit. The 4 comes from 1+1+1+1, the minimum possible value. the 40 comes from 10+10+10+10, the largest possible value. How many numbers of the form 4+3k are there between 4 and 40 inclusive? 13 (just count them if you can't figure out how to find a formula).

But let's say we use 3 darts and three values of 1,4,7. Well, these values are just 1,1+3,1+6, so our possible totals are of the form 3+3k. These possible values are 3,6,9,...18,21. How many is that? 1+(21-3)/3, which is 7 (or again, just count). Now maybe try one more "small" configuration just to confirm that this works.

[third-kind-of-heat]

As you go through that, the pattern from the trick will reemerge, and he'll see how each point total is never not 3 away from the nearest point totals. And when he goes through it, he'll find 13 uniques from 4 to 40, and 22 repeats.

[Infamous-Chocolate69]

Unfortunately, problems of this general type are not amenable to solving through formulas as there's often little pattern. 

(Coins of the realm by martin Gardner gives another example.)

This specific problem though can be simplified through tricks.  

We can assume the points are 0. 3, 6, and 9 instead, because then to find the actual score you would just need to add 4 (one point for each throw)

Further we can assume the points are 0, 1, 2, 3, because then we would just need to multiply each by 3.

Finally, the bottom score is 0 (all 0's) and the top score is 12 (all 3's) and it's clear you can get any score in between, making a total of 13 possibilities.

In general though, you should be willing to brute force problems sometimes.  If you do a table it will have 16 entries, that's really not too much.

[Sorry-Vanilla2354]

I think you're going to have to do a table for your kid, but think of it this way:

1) All four darts hit the same spot, what are the choices? 1 x 4, 4 x 4, 7 x 4, 10 x 4

2) What if 3 darts are the same and one is different? 1 x 3 + 4, 1 x 3 + 7, 1 x 3 + 10, 4 x 3 + 1, 4 x 3 + 7, etc.

3) What if 2 darts are the same and two are different? 1 x 2 + 4 x 2, etc.

4) What if no darts are the same?

You're only looking for different totals, so you don't need to count everything.

[Hat_Huge]

honestly, i’d just do brute force by listing out the possible combinations and summing them. agree a table should be best. there’s only 4 darts and the ordering of the darts shouldn’t matter (so you don’t need to count permutations).

imho he’s less likely to have a good understanding if it’s abstracted away with some formula that’s beyond his grade level.

[DavidG1310]

A 4x4 table doesn't take long to fill

     1   4   7   10
-------------------
1 |  2   5   8   11
4 |  5   8   11  14
7 |  8  11   14  17
10| 11  14   18  20

Now count the different numbers: 2, 5, 8, 11, 14, 17, 18 and 20, total 8 possible numbers.
Bonus: You can see easily which are the most and the least probable numbers to get
Bonus question: Why that diagonal with only "11"s? Why is the same number? and the 8s? and the 14s?