# Permutations and combinations

## Factorials

It is convenient to define n factorial, denoted as n! as the product of the first n positive integers. For example, 4! = 4 × 3 × 2 × 1 = 24. It is manifest that 1! = 1, and we define 0! = 1.

## Permutations

A permutation is an ordering or arrangement of objects. P(n,r) denotes the number of distict arrangements of r objects from n objects. For example P(5,2) = 20 because there are 20 ordered pairs from the letters abcde, viz. ab, ac, ad, ae, ba, bc, bd, be, ca, cb, cd, ce, da, db, dc, de, ea, eb, ec, ed. This number can be obtained as 5 × 4 = 20, because there are five choices for the first letter, and after that is removed, four choices remain for the second letter, You are chooosing the first letter and the second letter, hence this is an example of the multiplication (and) rule.

Note that the number of arrangements of n distinct objects is P(n,n) = n!.

The number of ways you can choose a president, vice-president, and secretary from a class of seven students is P(7,3) = 7 × 6 × 5 = 210. This can also be written a 7!/(7-3)!. In general P(n,r) = n!/(n-r)!; This can this can be interpreted as arranging all n objects, and then removing the order of the (n-r) objects which are not chosen by dividing by the number of ways to arrange them.

Exercise: If you have 7 distinct objects, how many permutations are there of all 7?, 6 of them?, five of them? four of them? three of them? two of them? one of them? zero of them? What does permuting one object or zero objects mean?

## Combinations

C(n,r) (which is read as n choose r) is the number of different unordered samples of size r which can be chosen from n distinct objects. For example, C(5,2) = 10 because ab, ac, ad, ae, bc, bd, be, cd, ce, de are the only pairs of letters which can be chosen from abcde (P(5,2) = 20 because each of these pairs can be ordered two ways as listed above). C(5,2) = 5!/((5-2)!×2!) which can be interpreted as arranging all 5 objects, and then removing order among the 3 which are not chosen and also among the 2 which are chosen. In general C(n,r) is equal to n!/((n-r)!×r!)

The classical notation for C(n,r) has n above r within parenthesis, but it is hard to do that in ascii, and it is not the notation of the text.

## Generalized combinations

Often one is interested in distinguishing some, but not all, of the individuals which are chosen. For example, consider the problem of selecting a president, vice-president, and a committee of three from seven individuals. This can be done by first selecting hte president and vice-president, and then choosing the committee ot three. This line of reasoning produces the algebra: P(7,2) × C(5,3) = (7!/5!) × (5!/(2! × 3!)) = 420. This can also be written as 7!/(1!1!3!2!) = 420 which can be interpreted as arranging all 7 individuals, then removing order among the presidents (there is only 1, 1! = 1), vice-presidents (there is only one), winners (there are three), and losers (there are 2). THis reasoning can also be extended to the question of how many arrangements are there of the letters in banana. There are 6!/(1!3!21) = 60; this can be interpreted as arranging all 6 letters, then removing the order among the b, the a's, and the n's.

Competencies:How many ways can a president,vice-president, and secretary be chosen from a class of 8? How many ways can a committee of three be chosen from a class of 8? How many ways can a president, vice-president, secrtary, and a committee of three be chosen from a class of 8?

How many arrangements are there of the letters in Mississippi?

Reflection:

Challenge:How many ways can you form 4 pairs of roommates from eight men?

May 2002