# Sunday Times Teaser 2907 – Combinatorial Cards

### by Susan Bricket

#### Published June 10 2018 (link)

On holiday once with lost luggage and trapped indoors, we decided to recreate our favourite card game. With limited resources, we used just seven cards and seven images (red heart, green tree etc) with three images on each card. Remarkably, just as in the game at home, every pair of cards had exactly one image in common. Labelling the seven images from 1 to 7, the cards were as follows:

{1,2,4} {2,3,6} {1,3,5} {1,6,7} {2,5,7} {3,4,7} {4,5,6}

Interestingly, each image appears on just three cards.

Our original set of cards at home has eight images per card, each image appears on just eight cards and again the total number of images is the same as the number of cards.

What is that number?

One Comment Leave one →
1. The solution to this teaser can be illustrated using the example given in the teaser. For seven cards there are $$C(7, 2) = 21$$ pairs of cards and we know that each pair shares only one image. We can now list the cards that share each of the images (1..7):
Here each row of three cards will yield $$C(3,2) = 3$$ pairs of cards. We can now see that the table contains a total of 21 cards which, among them, give 21 pairs of cards each of which share exactly one image. The table hence lists all the cards needed to make the 21 pairs needed for seven cards.
We can now generalise this for $$n$$ cards and $$n$$ images where each card has $$k$$ images and each image appears on $$k$$ cards. There will be $$n$$ rows and each row will contain $$k$$ cards from which we can make $$C(k,2)$$ pairs, giving the equation: $n C(k, 2) = C(n, 2)$ which immediately gives: $n=k(k – 1)+1$which, for $$k=8$$, gives a total of 57 cards.