(Discrete) Probability is essentially about counting. Counting the number of ways an outcome from a random experiment can happen. We need some tricks to help us. 😃
Permutations
A permutation of a set is an ordering of the set. If you have N elements in a set and you change the order you have another permutation.
Example 1.
Consider the set \( A=\{ 1,2,3 \} \). We can re-arrange the items in a set to form the following permutations
\begin{array}{cc}
(1,3,2) & (1,3,2) \\
(2,1,3) & (2,3,1) \\
(3,1,2) & (3,2,1) \\
\end{array}
We have \( 3! = 3*2*1 = 6 \) permutations (see next example for why this is the case)
The symbol \( N! \) (factorial) means \( N(N-1)…1 \)
Example 2.
Permutations of N different items.
If you have N different hats how many ways of ordering them are there? There are N ways to choose the first hat, N-1 ways to choose the second hat and so on.
We have in total $$ N(N-1)… 1 = N! $$ ways of ordering the hats.
Example 3.
Permutations with Repetition.
Suppose you N hats, K black and N-K red. Apart from the colour they are indistinguishable. How many ways of ordering them are there? We know from above N! if they are all different. However (N-K)! are indistinguishable (permutations of red hats) and K! are indistinguishable (permutations of black hats). Hence there are
$$\frac{N!}{K!(N-K)!}$$ orderings
Combinations
A combination of a set, \( A \), is a subset of \( A \), and unlike permutations above the order does not matter. An example will make this clearer.
Example 4.
Let’s suppose \( A=\{ 1,2,3 \} \). Then the following sets are all the combinations of \( A \)
\begin{array}{cc}
\{1,2\} & \{2,3\} & \{3,1\} \\
\{1\} & \{2\} & \{3\}\\
\{1,2,3\} & \emptyset & \\
\end{array}
You will notice there are 8 of them as each element in \( A \) can be chosen or not chosen for the combination ( 2 choices) and there are three elements in \( A \) all together so there are in total \( 2^3 =8 \) combinations.
To be more specific, subsets of size K are called K-combinations. Note that the sets \( \{1,2\}\) and \(\{2,1\}\) are the same 2-combination (subset) as changing the order of the elements does not create a new combination.
Lets concentrate on creating a K-combination of a set of N (distinct) elements. Note we don’t actually have to bother with the word distinct because by definition a mathematical set contains distinct elements.
Example 5.
K-combinations of an N-set
Suppose \( A=\{ 1,2,3,4 \} \) write down all the possible 3-combinations.
\begin{array}{cccc}
\{1,2,3\}&\{2,3,4\}&\{3,4,1\},&\{4,1,2\}
\end{array}
If we wanted to create all the permutations of subsets of \( A \) of size 3 we would have 4 choices for the first item, 3 choices for the second item and 2 choices for the third item i.e. 4*3*2 permutations in total. Each one of those sets above has 3! permutations as in example 1 since they contain 3 elements so that we know the number of combinations is \(\frac{4*3*2}{3!}\) or \(\frac{4!}{(4-3)!3!}\). In general the number of K-combinations from a set of size N is $$\frac{N}{(N-K)!K!}$$
\( \frac{N}{(N-K)!K!}\) is such an important formula it has its own notation.
$$ C(N,K) = \frac{N}{(N-K)!K!} $$
\( C(N,K) \) is sometimes also written as \( \binom{N}{K} \) and pronounced “N choose K”.