Setless Set (ITA Archived Puzzle, Analyzed)

Please refer to the setless set problem on this page.

Analysis

It can be easily deduced that the collection of 81 cards, contiains 27 SETs where each set contains the same cards.

Now in order to find the number of ways in which we can collect 20 cars such that it contains no set, we can use the following formula:
$$ N_{20\,cards\,without\,set} = N_{all\,possible\,collections\,of\,20\,cards} - N_{collection\,of\,20\,cards\,with\,atleast\,1\,set}$$


Find out set with different cards

Now if \(n_{same}\) is the number of sets with same cards, we already know that
$$n_{same} = 27$$

If we want to form a set that contains different cards, we can first pick an equal set and then pick a card from that set. Repeating this thrice will give us a set that contains different cards.

Thus total number of sets with different cards  \(n_{different}\) can be defined as 
$$n_{different} = (27 * 3) * (26 * 3) * (25 * 3) $$

Thus total number of possible sets \(n_{sets}\) can be defined as:
$$n_{sets} = n_{same} + n_{different}$$

Now we can easily find out the number of ways to pick k sets from \(n_{sets}\). A collection of 20 cards will contain a minimum of 1 set and a maximum of 6 sets. Thus number of ways to collect 'at the most' 6
sets can be defined as
$$n_{6\,sets} = \sum_{k = 1}^{k = 6} \binom {n_{sets}} {k}$$

Thus to find out N ways in which we can collect 20 cards that does not contain a set is
$$ N = \binom {81} {20} - n_{6\,sets}$$

Implementation

We can now write code that

• Computes \(n_{6\,sets}\)
• For a given j (j = 20 in the example), finds out the maximum number of ’k’ sets possible
• Computes the number of ways in which you can pick k sets that is \(n_{k\,sets}\)
• Computers the number of ways in which you can pick j cards from 81 cards
• Finally computes the number of ways in which you can pick j cards such that it contains no set.