1990 British 4-part TV mini-series surrounding the rise of (fictional) Conservative Party Chief Whip Francis Urquhart from a back-bencher to the seat of Prime Minister through his continual scheming and underhand dealings; through his varied connections he manufactures a crisis which forces the resignation of the prime-minister; his one failure comes when he kills a reporter (possibly) by accident.

It is hard to imagine that this series would have worked without Ian Richardson (as Urquhart) - a brilliant and formidable talent in shakespearean acting, he takes a script which otherwise would have left us with quite a cheesy and improbable Urquhart, and gives us one of the all-time best TV villains - his wry, smug asides to camera are a little distracting at first but become central to his attraction as a villain - he appears so far above those around him in intelligence and sense of humour that we have no choice but to hope he gets his way.

For students of Thatcher-era politics this show is a must-see.

The expression "house of cards" is also used to describe a fragile or tenuous organization, scheme, or situation. The idea being that like house of cards, any disturbance will cause the whole thing to collapse.

How many cards are there in a house of cards with r rows?

There are two ways we can approach this. One is simpler mathematically, but requires one to spot a special property of the construction of the house. The other is somewhat more challenging, but takes a more analytical approach in it's derivation. Of course, the equation constructed at the end is the same. I'll go through both methods and you can read either or both as you like.

The (slightly) inspired way

Lets look at a specific house of cards just to get the picture in our minds:

```       /\
_/__\_
/ \  / \
/___\/___\
/ \  / \  /\
/   \/   \/  \```

That's a house of cards for r = 3. You'll notice that the bottom row has no horizontal cards, while all the other rows do. These exceptions are bound to any formula more complicated, so for now we'll imagine that the bottom row also has these base cards. We can correct for this deliberate mistake later; simplication is a powerful technique used by real mathematicians when getting to grips with a new problem.

Here's our new, temporary house:

```       /\
_/__\_
/ \  / \
/___\/___\
/ \  / \  /\
/___\/___\/__\```

Now, in true mathematical style, we can reduce this to a previously solved problem. How? Well, we can treat each block of three cards (left slanter, right slanter and base card) as one "unit". And if we draw these units on their own...

```  .
. .
. . .```

A good old triangle! The formula for the nth triangular number is ½n(n+1), and a house of cards with r rows corresponds to a triangle of ½r(r+1) units, each of which contains 3 cards. So the formula of our slightly unorthodox house of cards is

```     3
C = --- · r(r+1),
2```

where C is the number of cards required.

Now we simply need to correct for our over-estimate in adding an extra row at the bottom. How many cards did we add? r of course! So we simply remove that quantity from our equation and we get the true number of cards, leaving us with the straight forward equation

```     3
C = --- · r(r+1) - r.
2```

The analytical way

Let's start by finding the number of cards in each row. However big the house of cards, the top row has the same number of cards, as does the next one down, and so on to the bottom of the house of cards. For this reason, we'll number the rows in this way:

```       /\        1
_/__\_
/ \  / \     2
/___\/___\
/ \  / \  /\   3
/   \/   \/  \```

We define a row as all the leaning cards plus the horizontal cards directly above it. Thus row 1 always has 2 cards, row 3 has 5 and so on. Let's make a quick table of the first few values of row R.

``` _________
| R  | CR |
|----|----|
| 1  |  2 |
| 2  |  5 |
| 3  |  8 |
| 4  | 11 |
|____|____|```

Now of course this method is relying on strictly analytical methods, so we're not allowed to spot the obvious pattern above. Instead, look at the house of cards. Each row contains R cards leaning right, R cards leaning left, and each of the R pairs of leaning cards has a horizontal card immediately above and to it's left, except for the right-most one. We could write this as the equation

``` CR = R + R + (R - 1)
= 3R - 1.```

So, in a house of cards with r rows, we need to add together the number of cards in each row from 1 up to and including r. In symbols,

```      r
__
\
C =  >  (3R - 1)
/__
R = 1```

If you're unfamiliar with the algebra of sum notation, you can read up on it here. Otherwise, the following steps should be self-explanatory and I have deliberately moved one step at a time.

```      r
\
C =  >  (3R - 1)
/
R = 1

r
\
=  >  (3R) - r
/
R = 1

r
\
= 3 · >  R   - r
/
R = 1

/ r + 1 \
= 3r| ------- | - r   [ Sum of the first N integers is ½n(n+1) ]
\   2   /

3
= --- · r(r+1) - r.
2```

Simplifying this final equation

We can do some straight-forward simplification of this equation, which is the same whatever method we used so far. This leaves with a much tidier equation to finish.

```     3
C = --- · r(r+1) - r  [Clear factor of r]
2

/ 3               \
= r |  - · (r + 1) - 1  |  [Multiply out those inner brackets]
\ 2               /

/ 3r   3      \
= r |  -- + -  -  1 |
\ 2    2      /

/ 3r   1 \
= r |  -- + -  |
\ 2    2 /

= ½r · (3r + 1)```

And there you have it, in it's simplest form. I do hope that was informative.

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