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Factorials and Permutations

Factorials

Many counting and probability calculations involve the product of a series of consecutive integers. A factorial, denoted as \(n!\), is the product of all positive integers from down to 1. You can use factorial notation to write such expressions more easily.

For any natural number n, the factorial is defined as the following:

\(n! = n \times (n - 1) \times (n - 2) \times (n - 3) \times ... \times 3 \times 2 \times 1\)

This is read as n factorial.

Note: \(1! = 1\), and \(0! = 1\) (by definition)

\(n\)! increases dramatically as n becomes larger, so you'll be using calculators and computer software to calculate the larger factorials.


Example

What is the result when \(4!\) is computed?

We can use the following formula to determine \(4!\):

\(n! = n \times (n - 1) \times (n - 2) \times (n - 3) \times ... \times 3 \times 2 \times 1\)

Next, we can substitute the pertinent values and solve:

\(4! = 4*3*2*1\)

\(4! = 24\)

Therefore, we can determine that \(4!\) is equal to 24.


Assume there are ten cups on a table, varying in height. In how many ways could the ten cups be arranged, if the tallest and shortest cup were placed beside each other.

We can use the following formula to determine the number of possible cup combinations:

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

Treat the tallest and shortest cups as a unit making nine items that are to be arranged. The two cups can be arranged in two ways.

Using these values, we can calculate the number of possible cup combinations as such:

\(9! * 2! = 725 760\)

Therefore we can determine you could arange the 10 cups in 725,760 unique ways.

Permutations

A permutation is an ordered arrangement of objects. The number of ways to arrange distinct objects in a sequence is given by \(n!\)

However, if we only select \(r\) objects out of \(n\), we use the permutation formula:

\( P(n, r) = \cfrac{n!}{(n - r)!}\)

This formula calculates the number of ways to arrange \(r\) objects chosen from a set of \(n\) objects, where order matters.

The total number of such permutations is denoted by:

\(P(n,r) = nPr\)

There are some special cases and properties for permutations:

  • If \(r = n\), then \(P(n,n) = n!\), meaning all objects are arranged
  • If \(r = 1\), then \(P(n,1) = n\), as there are ways to select one object
  • If \(n = r = 0\), then \(0! = 1\), so \(P(0,0) = 1\)

In how many ways can two students be selected and arranged in a row from a group of eight?

We use the factorial formula for determining the number of ways two students can be arranged to form a group of eight:

\( P(n, r) = \cfrac{n!}{(n - r)!}\)

We can then substitute the pertinent values and solve:

\(P(8, 2) = \cfrac{8!}{(8 - 2)!}\)

\(P(8, 2) = \cfrac{8!}{6!}\)

\(P(8, 2) = \cfrac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2\times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1}\)

\(P(8, 2) = 8 \times 7\)

\(P(8, 2) = 56\)

Therefore, we can determine there are 56 ways two students can be arrannged to form a group of eight.