Math Is Fun Forum

My assumption has been a sequence of 3-6-10-15-21 i.e.
a minimum of 3 for 5 numbers
a minimum of 6 for 6 numbers
a minimum of 10 for 7 numbers
a minimum of 15 for 8 numbers

This using a basis of differentials, e.g. for 3 the minimum is 1, for 4 it is 2 for 5 it is 3, therefore the minimum increases by an increment of 1 on top of the previous answer i.e.

a minimum of 1 for 3 numbers = 1 (1 = base figure)
a minimum of 2 for 4 numbers = 2 (1 + 1 = 2)
a minimum of 3 for 5 numbers = 3 (1 + 2 = 3)
a minimum of 6 for 6 numbers = 6 (1 + 2 + 3 = 6)
a minimum of 10 for 7 numbers = 10 (1 + 2 + 3 + 4 = 10)
a minimum of 15 for 8 numbers = 15 (1 + 2 + 3 + 4 + 5 = 15)

This is probably incorrect thinking, as there are more triples than quads for 4 numbers, the same for 5 numbers and less for 6 or more numbers.

Another thought I have had, is to assume that if 5 quads is the maximum coverage by any single triple, then the total of 70 quads divided by 5 is 14 and maybe that is the minimum. Again, this is probably incorrect, as for 6 numbers there are 15 quads and not all the triples closed 3 quads. Some closed 3, others closed 2.

I think you have taken me too literally, perhaps it could be reworded as "I think that this can be worked out using a mathematical formula, I don't know if it could I am just assuming it could."

The reason why I asked the question is because I thought that there would be a way of calculating this and maybe someone knew.

Yes please.

I see you have put some answers down, thank you for this. I can't help feeling 25 is a bit high, i.e. not the minimum.

I have come up with a lower number. I believe that all 70 quads are covered with the following 22 triples...

These triples each cover 5 quads...
1    2    4
1    3    8
1    6    7
2    3    5
2    7    8
3    4    6
4    5    7
5    6    8
This means that there are just 30 quads remaining that are not "closed".
       
These triples each cover 3 of the remaining "open" quads...
1    4    8
1    5    6
1    5    7
2    3    6
2    4    6
3    4    7
This now leaves just 12 that are "open".

These triples each cover 2 quads...
2    5    8
3    5    8
5    6    7
6    7    8

Meaning that there are just 4 quads left which are covered by....
1    2    3
1    2    6
2    3    4
3    4    5

What I would like to know is - "is this the minimum ?".

Thanks for your input so far, it has helped me to understand a bit more.

My apologies, it is difficult to try and describe because I don't really understand myself. Let me try again...

Combinations of 4 will be referred to as "quad".
Combinations of 3 will be referred to as "triple".
A point of clarification, we are looking for combinations and not permutations.

There are five number 1, 2, 3, 4, and 5. As we know, there are 5 different quad combinations. They are 1-2-3-4, 1-2-3-5, 1-2-4-5, 1-3-4-5 and 2-3-4-5. We also know that there are 10 triples possible i.e. 1-2-3, 1-2-4, 1-2-5, 1-3-4, 1-3-5, 1-4-5, 2-3-4, 2-3-5, 2-4-5, 3-4-5.

What I am looking for, is to find the minimum number of triples where the numbers appear in the quads. For example the triple 1-2-3 appears in 1-2-3-4 and 1-2-3-5. This leaves 3 other quads which do not feature the triple 1-2-3. For those quads that remain 1-4-5 appears in 1-2-4-5 and 1-3-4-5 and 3-4-5 covers 2-3-4-5.

This means that the minimum of triples required to appear in all the quads is  3.

8 numbers produces 70 quads and 56 triples.

The triple 1-2-3 appears in 5 quads, i.e. 1-2-3-4, 1-2-3-5, 1-2-3-6 1-2-3-7 and 1-2-3-8, leaving the other 55 triples to cover the remaining 65 quads but not all are required.

My question is - "what is the minimum number of triples required in order to appear in quads which are made up from 8 numbers ?"

Another example

I have done some more research and found that for 6 numbers there are 15 quads and there are just 6 triples that appear in all of the quads.

1-2-3 appears in 1-2-3-4, 1-2-3-5 and 1-2-3-6.
Of the 12 quads remaining, 1-5-6 appears in 1-2-5-6, 1-3-5-6 and 1-3-5-6.
Of the 9 quads remaining, 3-4-5 appears in 1-3-4-5, 1-3-4-5 and 13-4-5.
This now leaves 6 quads for which 1-2-4 appears in 1-2-4-5 and 1-2-4-6, 3-4-6 which appears in 1-3-4-6 and 2-3-4-6 and finally 2-5-6 which appears in the last 2, 2-3-5-6 and 2-4-5-6.

Unless I am wrong, it means that the minimum number of triples needed to appear in all 15 quads, is 6.

I hope this makes my puzzle easier to understand.