I need to find the solution of this pattern:

{a,b,c}, {c,d,e}, ?, ?, ?, {b,c,e}, {a,c,d}, {b,c,d}, {a,c,e}, ?

The solution should be something like this:

{a,b,c}, {c,d,e} {x,x,x} {x,x,x} {x,x,x},{b,c,e},{a,c,d}, {b,c,d}, {a,c,e}, {x,x,x}

where instead of the x you should replace letters.

Source: http://zagaza.ru/za548.htm


How about:

{a,b,c}, {c,d,e}, {a,b,d}, {b,d,e}, {a,b,e}, {b,c,e}, {a,c,d}, {b,c,d}, {a,c,e}, {a,d,e}.


We know there are $\binom53=10$ variants, and there are $10$ slots, so a one-to-one correspondence seems likely.

The lexical order is {a,b,c}, {a,b,d}, {a,b,e}, {a,c,d}, {a,c,e}, {a,d,e}, {b,c,d}, {b,c,e}, {b,d,e}, {c,d,e}.

Alternate indices cover 1,2,3,4,5 and 10,9,8,7,6, i.e. 1,10,2,9,3,8,4,7,5,6.

  • $\begingroup$ This is logical. I was going for Permutations instead. $\endgroup$ – ABcDexter Jan 6 at 18:34
  • $\begingroup$ thanks for answering,could you please explain me how you understood the order of those variants,i mean inside the x.x.x at the beginning there could even be C as first letter,please explain that to an heretical guy xD $\endgroup$ – alnesi Jan 6 at 18:55
  • $\begingroup$ do u have any other contact,like discord,steam even a twitter account that you use to communicate thanks in advace $\endgroup$ – alnesi Jan 6 at 19:28
  • $\begingroup$ @alnesi I think the explanation is already very good and clear. maybe you don't see how the lexical order is determined? Just take three letters, and order them as if they were words: "abc", "abd",... Once that is established, the index alternation should pretty easy. $\endgroup$ – Silly Freak Jan 6 at 19:31
  • $\begingroup$ oh thank you,i just wanted to know that :D .Really thank you $\endgroup$ – alnesi Jan 6 at 19:38

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