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Preview Combinations Calculator Widget

Combinations Calculator calculates the number of ways of selecting r outcomes from n possibilities when the order of the items chosen in the subset does not matter.

Combinations

6

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- The rules for using the combinations calculator
- The fundamental principle of counting
- Sample Spaces
- Combination
- Permutation
- The Difference between Combinations and Permutations

There are different strategies to determine the number of ways to choose objects from a given set in mathematics. In how many ways can we pick *r* outcomes from *n* possibilities? It depends on whether the order matters or not and can the values repeat or not.

The number of ways of picking r unordered outcomes from n possibilities is known as a combination and is written as C(n, r). It is also known as the binomial coefficient. This calculator enables you to calculate the combination of r objects from a set of n objects.

For a given set of objects, there is a certain number of ways to arrange or select some or all of them according to some order or specification. The calculator computes the number of ways of selecting r objects from a set of n objects without repetition and when the order does not matter. The calculator requires two inputs:

- n = number of distinct objects to choose from, and
- r = number of positions to fill.

An essential criterion for entering data into the combination calculator is that

$$0 ≤ r ≤ n$$

If you enter a number r that is larger than n, it will print a message

"Please enter 0 ≤ r ≤ n".

The Fundamental Counting Principle guides us in finding ways to accomplish different tasks. There are two fundamental rules of counting.

The first task can be done in m ways, and the second task can be done in n ways. If the tasks can not be done simultaneously, the number of possible ways can be counted as (m + n).

The first task can be done in m ways and the second task can be done in n ways. If both tasks can be done simultaneously, then there are (m × n) ways of performing them.

The cafeteria sells 3 kinds of pies and 4 kinds of drinks. Among them are apple pie, strawberry pie, and blueberry pie. And orange, grape, cherry and pineapple juice. Both drinks and pies are sold for $ 2. You only have $ 2 with you and not a cent more. So you have 3 + 4 = 7 opportunities to make some particular choice.

Suppose you want to count the number of ways to flip a coin and roll a die. The number of ways you can flip a coin is 2 as a coin has 2 faces. Likewise, there are 6 possible ways that you can roll a die. As you can do both the tasks simultaneously, there are then 2 × 6 = 12 ways that you can flip a coin and roll a die.

If you want to draw 2 cards from a deck of 52 cards without replacing them, then there are 52 ways to draw the first and 51 ways to draw the second. Therefore, the number of ways to draw two cards is 52 × 51 = 2,652.

A sample space is a listing of all the possible outcomes and is denoted by the capital letter *S*. The sample space for flipping a coin and rolling a die simultaneously are

*S = {{H,1}, {H,2}, {H,3}, {H,4}, {H,5}, {H,6}, {T,1}, {T,2}, {T,3}, {T,4}, {T,5}, {T,6}}*

There are twelve possible ways. The counting principles allow us to figure out the number of ways of experimenting without having to list them all out.

The number of possible ways of picking r non-repeating outcomes from n possibilities when the order is irrelevant is known as combination. The combination of objects is written as C (n, r). It is also known as the binomial coefficient. The combination formula is defined as

$$C(n,r)=\frac{n!}{r!(n-r)!}$$

The sign ! after a number or letter means that we are using the factorial of some number. For example, n! is the factorial of the number n - or the product of natural numbers from 1 to n. The factorial of number 2 is 1 × 2. The factorial of number 3 is 1 × 2 × 3. The factorial of number 4 is 1 × 2 × 3 × 4. The factorial of number 5 is 1 × 2 × 3 × 4 × 5 and so on. The factorial can only be calculated for non-negative integers.

An essential characteristic of calculating the combination using this formula is that the repetition of objects is not allowed, and the order of arrangement does not matter.

Suppose you have a set of four numbers

*{1, 2, 3, 4}*

In how many ways can we combine two elements from this set if the same element cannot be repeated in a pair?

If the order of the elements matters, we get groups formed by permutations:

*(1,2), (1,3), (1,4), (2,1), (2,3), (2,4), (3,1), (3,2), (3,4), (4,1), (4,2), (4,3)*

If the order does not matter - we get groups formed by combinations:

*(1,2), (1,3), (1,4), (2,3), (2,4), (3,4)*

There are 6 possible combinations. You can use the formula to find the number of all possible combinations. For this example, $n=4$, $r=2$. Therefore,

$$C(4,2)=\frac{4!}{2!(4-2)!}=\frac{4!}{2!2!}=\frac{4 × 3 × 2 × 1}{(2× 1)(2× 1)}=\frac{24}{4}=6$$

This is exactly what the Combinations Calculator calculates.

What are the combinations of letters A, B, C, and D in a group of 3? There are 24 possible permutations when the order is important. In combinatorial counting, the order is irrelevant. Therefore, only the first row is relevant, i.e., there are 4 possible combinations.

ABC | ABD | ACD | BCD |
---|---|---|---|

ACB | ADB | ADC | BDC |

BAC | BAD | CAD | CBD |

BCA | BDA | CDA | CDB |

CAB | DAB | DAC | DBC |

CBA | DBA | DCA | DCB |

Rather than listing all the possible arrangements, we can calculate the number of possible arrangements (in which the order is not important) using the combination formula above. Here, there are n=4 objects, and you are taking r=3 at a time. Therefore,

$$C\left(n,r\right)=C\left(4,3\right)=\frac{4!}{\left(4-3\right)!3!}=\frac{4!}{1!3!}=4$$

Permutation defines the number of ways to organize objects when the order of objects is important. The formula for permutation when selecting r objects from a list of n objects is as follows:

$$P\left(n,r\right)=\frac{n!}{\left(n-r\right)!}$$

The two main characteristics of calculating permutations using this formula are that object repetition is not allowed, and that the order of the objects is important.

Suppose there are 4 candidates in a job interview. The selection committee's task is to rank the candidates from 1 to 4. Here are the possibilities:

- 1st candidate - there are 4 ways to pick
- 2nd candidate - there are 3 ways to pick
- 3rd candidate - there are 2 ways to pick
- 4th candidate - there is only one way to pick

The product rule gives the total number of ways to pick, i.e. *4 × 3 × 2 × 1 = 24* which is the same as *4!*. Say the candidates are

{A, B, C, D}

The sample space of the problem, showing all the possible permutations, is shown below:

A in 1st place | B in 1st place | C in 1st place | D in 1st place |
---|---|---|---|

ABCD | BACD | CABD | DABC |

ABDC | BADC | CADB | DACB |

ACBD | BCAD | CBAD | DBAC |

ACDB | BCDA | CBDA | DBCA |

ADBC | BDAC | CDAB | DCAB |

ADCB | BDCA | CDBA | DCBA |

Rather than listing all the possible arrangements as shown in the table above, we can calculate the number of possible arrangements using the permutation formula. For the above example, there are n = 4 objects, and you take r = 4 elements at a time. Therefore,

$$P\left(n,r\right)=P\left(4,4\right)=\frac{4!}{\left(4-4\right)!}=\frac{4!}{0!}=24$$

The main difference between combinations and permutations is that in combinations the order of the elements is not important, while in permutations the order of the elements is important.

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