May 23

Permutations and combinations

by David Harper, CFA, FRM, CIPM

FRM | Quant |

In finance, sometimes the first question is, how many different outcomes are possible? To count up the number of outcomes or events, we can use permutations and combinations. (note that an event consists of none, one or more outcomes grouped depending on your purpose. When playing craps, the roll of seven is a single event achieved by way of several outcomes; six plus one, five plus two). In the binomial option pricing model, we can use a combination to quickly determine the number of possible price outcomes.

To calculate a permutation or combination, we only need the total number of objects (n = total objects) and the size of the "smaller" subgroup (r = how many objects we choose from the group.)

For example, let's say we invite five dinner guests (e.g., teacher, policewoman, artist, magician, and superman) but unfortunately our dinner table only seats three.

 

The number of permutations is the number of all possible configurations at the dinner table. In a permutation, we choose to seat three guests (r=3) among five total guests (n=5). The number of permutations is given by...

...and in this case, there are sixty permutations:

 

The number of combinations is not concerned with the order of any particular seating arrangements, only the group that is seated; if we seat superman, the artist, and the magician, this "group of three" is one combination regardless of where they sit at the table.

Combinations do not care about the sequence, only who's in the group. In a combination, we choose to seat three guests (r=3) among five total guests (n=5). The number of combinations is given by...

...and in this case, the number of combinations is ten:

 

You'll note the number of combinations must be less than the number of permutations. Further, the combination is equal to the permutation divided by r! (r factorial).

Comments

  1. Be the first to leave a comment!

Leave a Comment