Counting, permutations, and combinations

permutations example:

$A, B, C, D, E \longrightarrow 1, 2, 3, 4, 5$

Permutations: $5! = 5 * 4 * 3 * 2 * 1 = 120$

where $5!$stands for $5 factorial$.

$A, B, C, D, E \longrightarrow 1, 2, 3$

Permutations: $5 * 4 * 3 = 60 = \frac{5*4*3*2*1}{2*1} = \frac{5!}{2!} = \frac{5!}{(5-3)!}$

Factorial

$n! = n * (n-1)*(n-2)...1 \newline 3! = 3*2*1 \newline 2! = 2*1\newline 1!=1 \newline 0!=1$

$\text{Permutation formula:}\ \ _nP_k = \frac{n!}{(n-k)!}$

$\text{Combination formula:} \ \ _nC_k=\frac{\frac{n!}{(n-k)!}}{k!} = \frac{n!}{k!(n-k)!}=\binom{n}{k}$

example:

$A, B, C, D, E, F \text{(people)}\longrightarrow 1, 2, 3, 4 \text{(chairs)}$

$_6 C_4=\binom{6}{4} =\frac{6!}{4! (6-4)!} =\frac{61}{4!*2!}=\frac{6*5*4*3*2*1}{4*3*2*1\ \ *\ \ 2*1}= 15$

You will have 360 permutations, verses 15 combinations.

How many ways to choose 2 from $A, B, C, D$

$_4 C_2 = \binom{4}{2} = \frac{4*3}{2} = 6$

Another way to write this is $\binom{5}{3}$, or $5$ choose $3$, which is $10$.