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Enumeration Rules

How many different 5-girl teams I can make from a set of 10 girls, each girl's position matters in the choice. First we choose the center: there are 10 choices, then we choose Power forward: 9 choices, then we choose small forward: 8 choices, then we choose shooting guard: 7 choices, then the last(point guard) : 6 choices. The rule is to multiply the number of choices, I showed a way to see this in class with trees. the total number of different teams is :

\begin{displaymath}10 \times 9 \times 8 \times 7 \times 6=30240\end{displaymath}

1.
Multiplication Rule:
If k successive choices are to be made with exactly nj choices at each stage, then the total number of successive choices is

\begin{displaymath}n_1 \times n_2 \times \ldots n_k\end{displaymath}

2.
Number of orderings:
An ordering (or arrangement or permutation of k out of n) is a sequence of length k out of n choices with no duplications. The number of such arrangements is:

\begin{displaymath}n \times (n-1) \times (n-2) \times \ldots (n-k)\end{displaymath}

.
3.
Permutations: If the total number of possible choices is the same as the number of elements we want to pick, this is called a permutation, there are :

\begin{displaymath}n \times (n-1) \times (n-2) \times \ldots 1\end{displaymath}

permutations of n elements.

We have a special way of calling this number: ``factorial n'' = n! This number gets big very fast: 3!=6, 5!= 120, 10!=3628800, 20!= 2432902008176640000,
50!=30414093201713378043612608166064768844377641568960512000000000000.

4.
Combinations:
The number of ways of choosing k out of n objects is:

\begin{displaymath}\frac{n \times (n-1) \times (n-2) \times \ldots (n-k)}{
k \times (k-1) \times \ldots 1}= \frac{n!}{(n-k)!k!} =
{\binom{n}{ k}}\end{displaymath}

Property 1:

\begin{displaymath}{\binom{n} { k}}= {\binom{n-1} { k}} + {\binom{n-1} { k-1}}\end{displaymath}

Property 2:

\begin{displaymath}{\binom{n} { k}}={\binom{n} {n- k}}\end{displaymath}



 
next up previous index
Next: Birthday Problem 10/6 Up: Combinatorics Previous: Combinatorics
Susan Holmes
1998-12-07