Count Events:
Combinations, Permutations, and Factorials

Welcome to the Event Counter, a calculator that computes factorials, permutations, and combinations; as well as counts event multiples.

For help in using the calculator, read the Frequently-Asked Questions or review the Sample Problems.

Instructions: To find the answer to a frequently-asked question, simply click on the question. If none of the questions addresses your need, refer to Stat Trek's tutorial on the rules of counting or visit the Statistics Glossary. On-line help is just a mouse click away.

• What is a factorial?
• What is an event multiple?
• What is a permutation?
• What is a combination?
• What is the difference between a combination and a permutation?

What is a factorial?

In general, n objects can be arranged in n(n - 1)(n - 2) ... (3)(2)(1) ways. This product is represented by the symbol n!, which is called n factorial. By convention, 0! = 1.

Thus, 0! = 1; 2! = (2)(1) = 2; 3! = (3)(2)(1) = 6; 4! = (4)(3)(2)(1) = 24; 5! = (5)(4)(3)(2)(1) = 120; and so on.

Factorials can get very big, very fast. The term 170! is the largest factorial that the Event Counter can evaluate. The term 171! produces a result that is too large to be processed by this software; it is bigger than 10 to the 308th power.

For an example that computes a factorial, see Sample Problem 1.

What is an event multiple?

An event multiple refers to a grouping of two or more independent events. Many problems in statistics require a count of the number of possible event multiples.

For example, suppose one event is the outcome of a coin flip - heads or tails. Another event is the outcome from throwing a dart - hit or miss. These two events can be combined in four different ways - heads/hit, heads/miss, tails/hit, and tails/miss. Therefore, in this example, there are four possible event multiples.

For an example that counts event multiples, see Sample Problem 2.

What is a permutation?

A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement.

For example, suppose we have a set of three letters: A, B, and C. We might ask how many ways we can arrange 2 letters from that set. Each possible arrangement would be an example of a permutation. The complete list of possible permutations would be: AB, AC, BA, BC, CA, and CB.

When statisticians refer to permutations, they use a specific terminology. They describe permutations as n distinct objects taken r at a time. Translation: n refers to the number of objects from which the permutation is formed; and r refers to the number of objects used to form the permutation. Consider the example from the previous paragraph. The permutations were formed from 3 letters (A, B, and C), so n = 3; and each permutation consisted of 2 letters, so r = 2.

For an example that counts permutations, see Sample Problem 3.

What is a combination?

A combination is a selection of all or part of a set of objects, without regard to the order in which objects are selected.

For example, suppose we have a set of three letters: A, B, and C. We might ask how many ways we can select 2 letters from that set. Each possible selection would be an example of a combination. The complete list of possible selections would be: AB, AC, and BC.

When statisticians refer to combinations, they use a specific terminology. They describe combinations as n distinct objects taken r at a time. Translation: n refers to the number of objects from which the combination is formed; and r refers to the number of objects used to form the combination. Consider the example from the previous paragraph. The combinations were formed from 3 letters (A, B, and C), so n = 3; and each combination consisted of 2 letters, so r = 2.

Note that AB and BA are considered to be one combination, because the order in which objects are selected does not matter. This is the key distinction between a combination and a permutation. A combination focuses on the selection of objects without regard to the order in which they are selected. A permutation, in contrast, focuses on the arrangement of objects with regard to the order in which they are arranged.

For an example that counts the number of combinations, see Sample Problem 4.

What is the difference between a combination and a permutation?

The distinction between a combination and a permutation has to do with the sequence or order in which objects appear. A combination focuses on the selection of objects without regard to the order in which they are selected. A permutation, in contrast, focuses on the arrangement of objects with regard to the order in which they are arranged.

For example, consider the letters A and B. Using those letters, we can create two 2-letter permutations - AB and BA. Because order is important to a permutation, AB and BA are considered different permutations. However, AB and BA represent only one combination, because order is not important to a combination.

1. A standard deck of playing cards has 13 spades. How many ways can these 13 spades be arranged?
   
   Solution:
   
   The solution to this problem involves calculating a factorial. Since we want to know how 13 cards can be arranged, we need to compute the value for 13 factorial.
   
   
   13! = (1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13) = 6,227,020,800
   
   Note that the above calculation is a little cumbersome to compute by hand, but it can be easily computed using the Event Counter. To use the Event Counter, do the following:
   
   
   • Choose "Compute factorial" as the analytical goal.
   • Enter "13" for n.
   • Click the "Calculate" button.
   
   The answer, 6,227,020,800, is displayed in the "n Factorial" textbox.
   
   
2. On holidays, Chef Alice serves a dinner special consisting of a drink, an entree, and a dessert. Customers can choose from 5 drinks, 8 entrees, and 3 desserts. How many different meals can be created from the dinner special?
   
   Solution:
   
   The solution to this problem involves counting the number of event multiples. We know the following:
   
   
   • Each dinner consists of 3 independent events - a drink, an entree, and a dessert.
   • Customers can choose from 5 drinks, 8 entrees, and 3 desserts.
   
   The number of event multiples would be (5)(8)(3) = 120. Thus, 120 different meals can be created from the dinner special. To solve this problem using the Event Counter, do the following:
   
   
   • Choose "Count event multiples" as the analytical goal.
   • Enter "3" for "Number of independent events".
   • Enter "5" for "Number of ways event 1 can occur".
   • Enter "8" for "Number of ways event 2 can occur".
   • Enter "3" for "Number of ways event 3 can occur".
   • Click the "Calculate" button.
   
   The answer, 120, is displayed in the "Number of event multiples" textbox.
   
   
3. How many 3-digit numbers can be formed from the digits 1, 2, 3, 4, 5, 6, and 7, if each digit can be used only once?
   
   Solution:
   
   The solution to this problem involves counting the number of permutations of 7 distinct objects, taken 3 at a time. The number of permutations of n distinct objects, taken r at a time is
   
   
   _nP_r = n! / (n - r)!
   ₇P₃ = 7! / (7 - 3)! = 7! / 4! = (7)(6)(5) = 210
   
   Thus, 210 different 3-digit numbers can be formed from the digits 1, 2, 3, 4, 5, 6, and 7. To solve this problem using the Event Counter, do the following:
   
   
   • Choose "Count permutations" as the analytical goal.
   • Enter "7" for "Number of sample points in set ".
   • Enter "3" for "Number of sample points in each permutation".
   • Click the "Calculate" button.
   
   The answer, 210, is displayed in the "Number of permutations" textbox.
   
   
4. The Atlanta Braves are having a walk-on tryout camp for baseball players. Thirty players show up at camp, but the coaches can choose only four. How many ways can four players be chosen from the 30 that have shown up?
   
   Solution:
   
   The solution to this problem involves counting the number of combinations of 30 players, taken 4 at a time. The number of combinations of n distinct objects, taken r at a time is
   
   
   _nC_r = n! / r! (n - r)!
   ₃₀C₄ = 30! / 4!(30 - 4)! = 30! / 4! 26! = 27,405
   
   Thus, 27,405 different groupings of 4 players are possible. To solve this problem using the Event Counter, do the following:
   
   
   • Choose "Count combinations" as the analytical goal.
   • Enter "30" for "Number of sample points in set ".
   • Enter "5" for "Number of sample points in each combination".
   • Click the "Calculate" button.
   
   The answer, 27,405, is displayed in the "Number of combinations" textbox.