Suppose there are $3$ tennis players $\text{A}$, $\text{B}$, and $\text{C}$. A team consisting of $2$ players is to be formed. In how many ways can we do this? Is the team of $\text{A}$ and $\text{B}$ different from the team of $\text{B}$ and $\text{A}$? Unlike permutations, here the order is not important. In fact, there are only $3$ possible ways in which the team could be constructed and these are $\text{AB}$, $\text{BC}$, and $\text{CA}$.

Let’s understand what is combination and its types, and the formulas used to calculate with examples.

## What is Combination in Math?

In math, a combination is a way of selecting items from a collection where the order of selection does not matter. It is used in situations, as discussed in the beginning.

In smaller cases, it is possible to count the number of combinations, but for the cases which have a large number of groups of elements or sets, the possibility of a set of combinations is also higher. Therefore, a formula has been determined to find the possible selection of the number of items, which we will discuss later in the article.

Combinations are selections made by taking some or all of a number of objects, irrespective of their arrangements. The number of combinations of $n$ different things taken $r$ at a time, denoted by $^n \text{C}_r$.

### Examples of Combination** **

**Example 1:** There are $10$ players from where you want to pick a team of $3$ people. Picking a team of $3$ people from a group of $10$ is represented as $^{10} \text{C}_3$.

**Example 2:** There are $12$ desserts available on the menu in a restaurant and you want to select $4$ desserts for yourself and your friends. Choosing $4$ desserts from a menu of $12$ is represented as $^{12} \text{C}_4$.

**Example 3:** Make yourself a cup of coffee and observe the sequence in which you add milk, water, coffee, and sugar. Do you think your coffee will taste different if you add the ingredients in a different order to the cup? Not really. That’s because your coffee is a combination of its ingredients. Here the total number of combinations is $^4 \text{C}_4$, i.e., selecting $4$ items from a group of $4$ items.

**Example 4:** Your question paper mentions you can solve any two out of five questions in a section. Here the total number of items is $5$ and the number of items per selection is $2$, and the total number of combinations is $^5 \text{C}_2$.

## Combination Formula

The combination of $4$ objects taken $3$ at a time is the same as the number of subgroups of $3$ objects taken from $4$ objects. For example, given three fruits say an apple, an orange, and a banana, three combinations of two can be drawn from these sets (an apple and a banana), (an apple and an orange), or (a banana and an orange).

More formally, a $r$-combination of a set is a subset of $r$ distinct elements of $\text{S}$. If the set has $n$ elements, the number of $r$-combinations is equal to the binomial coefficient $^n \text{C}_r = \frac{(n)(n-1)(n-2)….(n-r+1)}{(r-1)(r-2) … 1}$

which can be written as $^n \text{C}_r = \frac{n!}{r!(n-r)!}$, when $n \gt r$

$^n \text{C}_k = 0$ , when $n \lt k$

where $n =$ distinct object to choose from

$\text{C} =$ Combination

$k =$ spaces to fill

### Examples on Combination Formula

**Example 1:** Find the value of $^5 \text{C}_2$

Here $n = 5$, and $r = 2$

According to combination formula: $^n \text{C}_r = \frac{n!}{r!(n-r)!}$

Therefore, $^5 \text{C}_2 = \frac{5!}{2!(5-2)!} = \frac{5!}{2! \times3!} = \frac{5 \times 4 \times 3!}{2! \times3!} = \frac{5 \times 4 }{2! } = \frac{5 \times 4 }{2 \times 1 } = 10$.

**Example 2:** If $^n \text{C}_2 = ^n \text{C}_7$, find $n$.

$^n \text{C}_r = \frac{n!}{r!(n-r)!}$

Therefore, $^n \text{C}_2 = \frac{n!}{2!(n-2)!}$

And, $^n \text{C}_7 = \frac{n!}{7!(n-7)!}$

Therefore, $\frac{n!}{2!(n-2)!} = \frac{n!}{7!(n-7)!} => \frac{1}{2!(n-2)!} = \frac{1}{7!(n-7)!}$

$ => 2!(n-2)! = 7!(n-7)! => \frac{(n-2)!}{(n-7)!} = \frac{7!}{2!}$

$=> \frac{(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)!}{(n-7)!} = \frac{7 \times 6 \times 5 \times \times 4 \times 3 \times 2!}{2!}$

$=> (n-2)(n-3)(n-4)(n-5)(n-6) = 7 \times 6 \times 5 \times \times 4 \times 3 $

$=>n – 2 = 7$, $=>n – 3 = 5$, $=>n – 4 = 5$, $=>n – 5 = 4$, and $=>n – 6 = 3$

$=> n = 9$

**Example 3:** A committee of $3$ persons is to be constituted from a group of $2$ men and $3$ women. In how many ways can this be done? How many of these committees would consist of $1$ men and $2$ women?

The number of men = $2$

The number of women = $3$

Total number of people = $n = 2 + 3 = 5$

Number of people in a committee = $r = 3$

a) The number of ways a committee of $3$ people from a group of $5$ people = $^5 \text{C}_3 = \frac{5!}{3!(5-3)!} = \frac{5!}{3! \times 2!} = \frac{5 \times 4 \times 3!}{3! \times 2 \times 1} = \frac{5 \times 4}{2 \times 1} = 10$.

b) Number of ways a committee of $1$ men and $2$ women.

The total number of men = $2$

The number of men to be included in a committee = $1$

This can be done in $^2 \text{C}_1$ ways

$^2 \text{C}_1 = \frac{2!}{1! \times (2 – 1)!} = \frac{2!}{1! \times 1!} = 2$

The total number of women = $3$

The number of women to be included in a committee = $2$

This can be done in $^3 \text{C}_2$ ways

$^3 \text{C}_2 = \frac{3!}{2! \times (3 – 2)!} = \frac{3!}{2! \times 1!} = 3$

Therefore, the number of ways a committee of $1$ men and $2$ women from a group of $2$ men and $3$ women can be formed in $2 \times 3 = 6$ ways.

**Example 4:** What is the number of ways of choosing $4$ cards from a pack of $52$ playing cards? In how many of these

a) four cards are of the same suit,

b) four cards belong to four different suits,

c) are face cards,

The number of cards in a deck of cards $n = 52$

The number of cards chosen from the deck $r = 4$

The number of ways $4$ cards can be chosen from a deck of $52$ cards = $^52 \text{C}_4 = \frac{52!}{4! \times (52 – 4)!} = \frac{52!}{4! \times 48!} = \frac{52 \times 51 \times 50 \times 49 \times 48!}{4 \times 3 \times 2 \times 1 \times 48!} = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1} = 270725$ ways

a) There are four suits: diamond, club, spade, heart and there are $13$ cards of each suit.

Therefore, there are $^{13} \text{C}_4$ ways of choosing $4$ diamonds.

Similarly, there are $^{13} \text{C}_4$ ways of choosing $4$ clubs

$^{13} \text{C}_4$ ways of choosing 4 spades

And $^{13} \text{C}_4$ ways of choosing 4 hearts

Therefore, The required number of ways = $^{13} \text{C}_4 + ^{13} \text{C}_4 + ^{13} \text{C}_4 + ^{13} \text{C}_4 = 4 \times ^{13} \text{C}_4 = 4 \times \frac{13!}{4! \times 9!} = 4 \times \frac{13 \times 12 \times 11 \times 10 \times 9!}{4 \times 3 \times 2 \times 1 \times 9!} = 2860$.

b) There are $13$ cards in each suit.

Therefore, there are $^{13} \text{C}_1$ ways of choosing $1$ card from $13$ cards of diamond, $^{13} \text{C}_1$ ways of choosing $1$ card from $13$ cards of hearts, $^{13} \text{C}_1$ ways of choosing $1$ card from $13$ cards of clubs, $^{13} \text{C}_1$ ways of choosing $1$ card from $13$ cards of spades.

Hence, by multiplication principle, the required number of ways = $^{13} \text{C}_1 \times ^{13} \text{C}_1 \times ^{13} \text{C}_1 \times ^{13} \text{C}_1$

$^{13} \text{C}_1 = \frac{13!}{1! \times 12!} = \frac{13 \times 12!}{1! \times 12!} = 13$

Therefore, $^{13} \text{C}_1 \times ^{13} \text{C}_1 \times ^{13} \text{C}_1 \times ^{13} \text{C}_1 = 13 \times 13 \times 13 \times 13 = 28561$ ways.

c) There are $12$ face cards and $4$ are to be selected out of these $12$ cards. This can be done in $^{12} \text{C}_4$ ways.

Therefore, the required number of ways = $\frac{12!}{4! \times (12 – 4)!} = \frac{12!}{4! \times 8!} = \frac{12 \times 11 \times 10 \times 9 \times 8!}{4 \times 3 \times 2 \times 1 \times 8!} = 495$.

## Relation between Permutation and Combination

The combination is a type of permutation where the order of the selection is not considered. Hence, the count of permutations is always more than the number of combinations. This is the basic difference between permutation and combination.

The relation between permutation and combination is $^n \text{P}_r = ^n \text{C}_r \times r!$

It means corresponding to each combination of $_n \text{C}_r$, there are $r!$ permutations because $r$ objects in every combination can be rearranged in $r!$ ways.

Let’s check the relation.

$^n \text{C}_r \times r! = \frac{n!}{r!(n-r)!} \times r! = \frac{n!}{(n-r)!} = ^n \text{P}_r$.

## Important Relation of Combinations

Another important relation of combinations is $^n \text{C}_r + ^n \text{C}_{r-1} = ^{n+1} \text{C}_r$.

Let’s prove the relation.

$^n \text{C}_r + ^n \text{C}_{r-1} = \frac{n!}{r! (n – r)!} + \frac{n!}{(r – 1)!(n – (r – 1))!}$

$ = \frac{n!}{r! (n – r)!} + \frac{n!}{(r – 1)!(n – r + 1)!}$

$= \frac{n!}{r(r – 1)! (n – r)!} + \frac{n!}{(r – 1)!(n – r + 1)(n – r)!}$

$= \frac{n!}{(r – 1)!(n – r)!} \times \left(\frac{1}{r} + \frac{1}{n – r + 1} \right)$

$ = \frac{n!}{(r – 1)!(n – r)!} \times \left(\frac{n + 1}{r(n – r + 1)} \right)$

$ = \frac{(n + 1)!}{r! (n + 1 – r)!} = ^{n + 1} \text{C}_r$.

## Difference Between Combination and Permutation

The following are the major differences between combination and permutation

Combination | Permutation |

The combination means the selection of objects, in which the order of selection does not matter. | The permutation means the selection of objects, where the order of selection matters. |

It is the selection of $r$ objects taken out of $n$ objects irrespective of the object arrangement. | It is the arrangement of $r$ objects taken out of $n$ objects |

The formula for combination is $^{n} \text{C}_{r} = \frac{n!}{r!(n – r)!}$ | The formula for permutation is $^{n} \text{P}_{r} = \frac{n!}{(n – r)!}$ |

## Practice Problems

- A box contains two white balls, three black balls, and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?
- Find the number of ways in which $4$ letters can be selected from the word EXAMINATION.
- How many triangles can be formed by joining $15$ points on the plane, in which no line joins any three points?
- A committee of $7$ members is to be chosen from $6$ artists, $4$ singers, and $5$ writers. In how many ways can this be done if in the committee there must be at least one member from each group and at least $3$ artists?
- How many different combinations of management can there be to fill the positions of president, vice president, and treasurer of a football club knowing that there are $12$ eligible candidates?
- How many combinations can the seven colors of the rainbow be arranged into groups of three colors each?
- How many different five-digit numbers can be formed with only odd-numbered digits? How many of these numbers are greater than $70,000$?
- $10$ people exchange greetings at a business meeting. How many greetings are exchanged if everyone greets each other once?
- How many lottery tickets must be purchased to complete all possible combinations of six numbers, each with a possibility of being from $1$ to $49$?
- Sarla has $8$ colored pencils that are all unique. She wants to pick three colored pencils from her collection and give them to her younger sister. How many different combinations of colored pencils can Sarla make from $8$ pencils?

## FAQs

### What are examples of combinations?

Some of examples of combinations are

a) There are $10$ players from which you want to pick a team of $3$ people. Picking a team of $3$ people from a group of $10$ is represented as $^{10} \text{C}_3$.

b) There are $12$ desserts available on the menu in a restaurant and you want to select $4$ desserts for yourself and your friends. Choosing $4$ desserts from a menu of $12$ is represented as $^{12} \text{C}_4$.

c) Your question paper mentions you can solve any two out of five questions in a section. Here the total number of items is $5$ and the number of items per selection is $2$, and the total number of combinations is $^5 \text{C}_2$.

### What do we mean by combinations?

A combination is a mathematical concept that determines the number of possible arrangements in a collection of items where the order of the selection does not matter. Unlike permutations, in combinations, you can select the items in any order.

### How do I calculate the number of combinations?

Combinations are a way to calculate the total outcomes of an event where the order of the outcomes does not matter. To calculate combinations, we use the formula $^n \text{C}_r = \frac{n!}{r!(n-r)!}$, where $n$ is the total number of items, and $r$ is the number of items being chosen at a time.

### How do you express combinations?

The number of combinations of $n$ objects taken $r$ at a time is determined by the following formula $^n \text{C}_r = \frac{n!}{r!(n-r)!}$.

### What is the rule of combination?

The basic rule of combination is **the order doesn’t matter**. If the order does matter, it is a permutation. A permutation is an ordered combination.

## Conclusion

A combination is a mathematical concept that determines the number of possible arrangements in a collection of items where the order of the selection does not matter. Unlike permutations, in combinations, you can select the items in any order. The formula used to select $r$ items from a group of $n$ items is given by $^n \text{C}_r = \frac{n!}{r!(n-r)!}$.