Combinations

Suppose you are walking into an ice cream shop to buy a $3$ scooped ice cream. There are $12$ flavours of ice cream available, and you need to choose $3$. How many ways can you make your choice? $3$ ways? $12$ ways? $36$ ways? No... It's much more than that. The concept of combinations helps in answering such sort of questions.

What are Combinations?

A combination is a mathematical technique for determining the number of possible arrangements in a collection of items where the selection order does not matter. We can pick the items in any order in combinations. Combinations are often confused with permutations. The order of the chosen items matters in permutations. In permutations of three letters, $abc, bca$, and $cab$ are different. But in combinations, all three are the same.

More precisely,
The order doesn't matter in a Combination.
The order does matter in a Permutation.

The number of possible combinations or selections from a pool of $n$ objects selecting only a few objects without any repetition is given by

${}^{\mathbf{n}}\mathit{C}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{r}\mathbf{!}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{r}\mathbf{)}\mathbf{!}}$

• $n$ is the total number of objects
• $r$ is the number of selected objects
• $!$ is the factorial notation

The factorial of any number $n$ is the product of all positive integers less or equal to $n$.
For example, $4!=4\times 3\times 2\times 1$.

Number of Combinations

In our ice cream shop example,
The number of available choices is denoted by $n=12$.
The number of flavours we choose denotes the $r$, and here, $r=3$.
So, the number of combinations of $3$ scooped ice creams that we can make is given by ${}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}=\frac{\mathrm{n}!}{\mathrm{r}!(\mathrm{n}-\mathrm{r})!}$
Where $r$ is a non-negative number that is less than or equal to $n$. This is also called the null formula.
So, the number of possible ice cream combinations

null
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We can manually count the number of combinations for smaller sets of objects. When done manually, there is a greater chance of error as the set size grows. The combination formula then enters the picture.

Representation of Combinations

We know that combinations are represented as null. The other common representations of null are:

• $\mathbf{C}\mathbf{(}\mathbf{n}\mathbf{,}\mathbf{r}\mathbf{)}$
• ${}_{\mathbf{r}}{}^{\mathbf{n}}\mathbf{C}$
• ${}_{\mathbf{n}}\mathbf{C}_{\mathbf{r}}$
• $$\begin{gathered} \left(n \\ r\right) \end{gathered}$$

The last representation $$\begin{gathered} \left(n \\ r\right) \end{gathered}$$ is also called the binomial coefficient.

Relation between Permutations and Combinations

We know that the permutations are given by null and combinations are given by the formula null, where r is the number of objects taken from a collection of n objects.

The formula for Combinations,


Hence, we can say that permutations are always greater than combinations for the same values of $r$ and $n$.

Some Special Cases

• When $r=n$, the number of combinations is given by,

${}^{\mathbf{n}}\mathbf{C}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{n}\mathbf{!}\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{n}\mathbf{)}\mathbf{!}}$
$\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{n}\mathbf{!}\mathbf{\times }\mathbf{0}\mathbf{!}}\mathbf{=}\mathbf{1}$
null

• The number of combinations to choose $0$ objects from a collection of $n$ objects is $1$. This is because combinations are different ways of selecting some or all the objects in the collection. Selection of nothing means we leave behind all the objects, and there is only one way of doing it.

When $r=0$, the number of combinations is given by,
null


Selecting $r$ objects from a collection of $n$ objects is the same as choosing $n-r$ objects.
If $r=n-r$, then,
null

null


If null , then, $\mathit{a}\mathbf{=}\mathit{b}$
Since null , we can say that $\mathit{b}\mathbf{=}\mathit{n}\mathbf{-}\mathit{b}$.
$$\begin{gathered} \mathbf{\therefore }\mathit{a}\mathbf{=}\mathit{n}\mathbf{-}\mathit{b}\mathbf{ } \\ \mathbf{\Rightarrow }\mathit{n}\mathbf{=}\mathit{a}\mathbf{+}\mathit{b} \end{gathered}$$

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Solved Examples

Q1. Akhil was buying three roses for his friend from a flower shop. There are roses in $\mathbf{6}$ colours. How many ways can he pick $\mathbf{3}$ roses trio out of total $\mathbf{6}$ roses?

Ans: To select $r$ objects from a set of $n$, null
Here, $n=6$ and $r=3$

null

So, he can have $20$ combinations of the rose trio when he chooses $3$ roses at a time from $6$.

Q2. How many $3-$letter combinations can be made using the letters of the word 'EXPERTSEDGE'?

Ans: The number of letters in the given word $=11$
Number of letters in the required combinations $=3$
Number of combinations of $3-$letter words null
So, we can make $165$ three-letter combinations can using the letters of the word 'EXPERTSEDGE'.

Q3. There were $66$ handshakes that happened at a party. Everybody shook hands with everybody else, and no one was left. How many people were there at the party?

Ans: Total number of handshakes, null 
Number of hands in a shake, $\mathrm{r}=2$
$\therefore$ The total number of people can be calculated using

null

The number of people is never negative, and hence the only possible value is $n=12$.

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