What is the math behind the game Spot It? I just purchased the game Spot It. As per this site, the structure of the game is as follows:

Game has 55 round playing cards. Each card has eight randomly placed symbols. There are a total of 50 different symbols through the deck. The most fascinating feature of this game is any two cards selected will always have ONE (and only one) matching symbol to be found on both cards.

Is there a formula you can use to create a derivative of this game with different numbers of symbols displayed on each card.
Assuming the following variables:


*

*S = total number of symbols

*C = total number of cards

*N = number of symbols per card


Can you mathematically demonstrate the minimum number of cards (C) and symbols (S) you need based on the number of symbols per card (N)?

$n^2 -n + 1$ where $n$ is the number of images.
This is the simplest formula to arrive at the number of both individual symbols and total number of cards required to display them (these are the same).
I derived this formula logically but not necessarily mathematically as follows:
I picked a random card and focused on a single image. Assuming eight images per card as are found in this game, this image can only be found $8$ times, once on the card you're holding and $7$ more times.
The same holds true for the next image. It can only appear $8$ times if it to remain unique - once on the card you are holding and once over each of $7$ more cards.
I noticed the trend. Each image appears once on the card you're holding and requires $7$ more cards. So, you need the 1 card you are holding and 7 more per image. Mathematically, I guess that's: 
$1 \text{card} + (7\text{cards}\times 8\text{images})$. That's $1+(7\times8)$ or $1+56 = 57.$
Logical, so far.
Then, I ran the same logic and considered a card with only $4$ images. Each card would require one base card and $3$ additional cards per image. Mathematically, that would be $1+ (3x4)$. That's $1+12$ or $13$ cards.
Then, I tried to tie these observations together. I asked myself "Is there a formula that would arrive at the right answer no matter the number of images?" The answer is yes. 
I remembered that in the examples above I started with 1 card then added (one less than the number of images) $\times$ (the number of images). That's $1+ (n-1)(n)$ if $n$ is the number of images. Then I just kinda rearranged a little:
$$\begin{eqnarray*}1+ (n-1)(n)  \\
1+ (n)(n) - n  \\
1+ n^2 - n  \\
n^2 - n + 1  
\end{eqnarray*}$$
I tested it and it works out every time. I was very happy before I got yelled at by my wife for taking so long on the computer.
