How many combinations of 2 with 8 numbers

The calculation uses the binomial coefficient: $$ C_n^k = \binom{n}{k} = \frac{n!}{k!(n-k)!} $$

3 choose 2 = 3 combinations (1,2)(1,3)(2,3) 4 choose 2 = 6 combinations (1,2)(1,3)(1,4)(2,3)(2,4)(3,4) 5 choose 2 = 10 combinations (1,2)(1,3)(1,4)(1,5)(2,3)(2,4)(2,5)(3,4)(3,5)(4,5) 6 choose 2 = 15 combinations (1,2)(1,3)(1,4)(1,5)(1,6)(2,3)(2,4)(2,5)(2,6)(3,4)(3,5)(3,6)(4,5)(4,6)(5,6) 7 choose 2 = 21 combinations (1,2)(1,3)(1,4)(1,5)(1,6)(1,7)(2,3)(2,4)(2,5)(2,6)(2,7)(3,4)(3,5)(3,6)(3,7)(4,5)(4,6)(4,7)(5,6)(5,7)(6,7) 8 choose 2 = 28 combinations (1,2)(1,3)(1,4)(1,5)(1,6)(1,7)(1,8)(2,3)(2,4)(2,5)(2,6)(2,7)(2,8)(3,4)(3,5)(3,6)(3,7)(3,8)(4,5)(4,6)(4,7)(4,8)(5,6)(5,7)(5,8)(6,7)(6,8)(7,8) 9 choose 2 = 36 combinations (1,2)(1,3)(1,4)(1,5)(1,6)(1,7)(1,8)(1,9)(2,3)(2,4)(2,5)(2,6)(2,7)(2,8)(2,9)(3,4)(3,5)(3,6)(3,7)(3,8)(3,9)(4,5)(4,6)(4,7)(4,8)(4,9)(5,6)(5,7)(5,8)(5,9)(6,7)(6,8)(6,9)(7,8)(7,9)(8,9) 4 choose 3 = 4 combinations (1,2,3)(1,2,4)(1,3,4)(2,3,4) 5 choose 3 = 10 combinations (1,2,3)(1,2,4)(1,2,5)(1,3,4)(1,3,5)(1,4,5)(2,3,4)(2,3,5)(2,4,5)(3,4,5) 6 choose 3 = 20 combinations (1,2,3)(1,2,4)(1,2,5)(1,2,6)(1,3,4)(1,3,5)(1,3,6)(1,4,5)(1,4,6)(1,5,6)(2,3,4)(2,3,5)(2,3,6)(2,4,5)(2,4,6)(2,5,6)(3,4,5)(3,4,6)(3,5,6)(4,5,6) 7 choose 3 = 35 combinations (1,2,3)(1,2,4)(1,2,5)(1,2,6)(1,2,7)(1,3,4)(1,3,5)(1,3,6)(1,3,7)(1,4,5)(1,4,6)(1,4,7)(1,5,6)(1,5,7)(1,6,7)(2,3,4)(2,3,5)(2,3,6)(2,3,7)(2,4,5)(2,4,6)(2,4,7)(2,5,6)(2,5,7)(2,6,7)(3,4,5)(3,4,6)(3,4,7)(3,5,6)(3,5,7)(3,6,7)(4,5,6)(4,5,7)(4,6,7)(5,6,7) 5 choose 4 = 5 combinations (1,2,3,4)(1,2,3,5)(1,2,4,5)(1,3,4,5)(2,3,4,5) 6 choose 4 = 15 combinations (1,2,3,4)(1,2,3,5)(1,2,3,6)(1,2,4,5)(1,2,4,6)(1,2,5,6)(1,3,4,5)(1,3,4,6)(1,3,5,6)(1,4,5,6)(2,3,4,5)(2,3,4,6)(2,3,5,6)(2,4,5,6)(3,4,5,6) 7 choose 4 = 35 combinations (1,2,3,4)(1,2,3,5)(1,2,3,6)(1,2,3,7)(1,2,4,5)(1,2,4,6)(1,2,4,7)(1,2,5,6)(1,2,5,7)(1,2,6,7)(1,3,4,5)(1,3,4,6)(1,3,4,7)(1,3,5,6)(1,3,5,7)(1,3,6,7)(1,4,5,6)(1,4,5,7)(1,4,6,7)(1,5,6,7)(2,3,4,5)(2,3,4,6)(2,3,4,7)(2,3,5,6)(2,3,5,7)(2,3,6,7)(2,4,5,6)(2,4,5,7)(2,4,6,7)(2,5,6,7)(3,4,5,6)(3,4,5,7)(3,4,6,7)(3,5,6,7)(4,5,6,7) 6 choose 5 = 6 combinations (1,2,3,4,5)(1,2,3,4,6)(1,2,3,5,6)(1,2,4,5,6)(1,3,4,5,6)(2,3,4,5,6) 7 choose 5 = 21 combinations (1,2,3,4,5)(1,2,3,4,6)(1,2,3,4,7)(1,2,3,5,6)(1,2,3,5,7)(1,2,3,6,7)(1,2,4,5,6)(1,2,4,5,7)(1,2,4,6,7)(1,2,5,6,7)(1,3,4,5,6)(1,3,4,5,7)(1,3,4,6,7)(1,3,5,6,7)(1,4,5,6,7)(2,3,4,5,6)(2,3,4,5,7)(2,3,4,6,7)(2,3,5,6,7)(2,4,5,6,7)(3,4,5,6,7)

Find the number of combinations and permutations when you choose a subset of r elements from a set of n elements. For help in using the calculator, read the Frequently-Asked Questions or review the Sample Problem.

Frequently-Asked Questions

Instructions: To find the answer to a frequently-asked question, simply click on the question.

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.

Note that AB and BA are considered to be two permutations, because the order in which objects are selected matters. 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. Bottom line: AB and BA represent two permutations, but only one combination.

For an example that counts permutations, see Sample Problem 1.

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. Bottom line: AB and BA represent two permutations, but only one combination.

For an example that counts the number of combinations, see Sample Problem 2.

How do you count the number of combinations?

A combination is a selection of all or part of a set of objects, without regard to the order in which they were selected. This means that XYZ is considered the same combination as ZYX.

The number of combinations of r objects that can be selected from a set of n objects is denoted by nCr. And the formula for computing that number is:

nCr = n(n - 1)(n - 2) ... (n - r + 1)/r! = n! / r!(n - r)!

Note: In the formula above, n! refers to n factorial, where n! is equal to n(n-1)(n-2) ... (3)(2)(1).

How do you count the number of permutations?

A permutation is a selection of all or part of a set of objects, with regard to the order in which they were selected. This means that XYZ is considered a different permutation than ZYX.

The number of permutations of r objects that can be selected from a set of n objects is denoted by nPr. And the formula for computing that number is:

nPr = n(n - 1)(n - 2) ... (n - r + 1) = n! / (n - r)!

Note: In the formula above, n! refers to n factorial, where n! is equal to n(n-1)(n-2) ... (3)(2)(1).

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.

What is E Notation?

The Combinations and Permutations Calculator uses E notation to express very large numbers. E notation is a way to write numbers that are too large or too small to be concisely written in a decimal format.

With E notation, the letter E represents "times ten raised to the power of". Here is an example of a number written using E notation:

3.02E+12 = 3.02 * 1012 = 3,020,000,000,000

How accurate is this calculator?

When the number of combinations or permutations is displayed as an ordinary integer, the result is exact. When the number of combinations or permutations is displayed using E notation, the result is not exact; but it is a very good approximation, accurate to 16 decimals.

Sample Problem

1. 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:

   nPr = 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 Combination and Permutation Calculator, do the following:

   • Enter "3" for "Subset size".
   • Enter "7" for "Set size".
   • Click the "Calculate" button.

   The answer, 210, is displayed in the "Permutations" textbox, as shown below.

1. 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:

   nCr = 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 Combination and Permutation Calculator, do the following:

   • Enter "4" for "Subset size".
   • Enter "30" for "Set size".
   • Click the "Calculate" button.

   The answer, 27,405, is displayed in the "Combinations" textbox, as shown below.

How many pairs can you make with 8 numbers?

What is the value of 8 Combination 2?

How many different combinations of 2 are there?

How many permutations of two digit numbers having 8 can be made?