# Combinations

Hey kids! Did you know that if you have a class of 15 students, then the total number of ways of making any 4 students out of 15, stand in a circle will be different than making them stand in a row? This is one of the properties of combinations. In this mini-lesson, we will be learning all combinations, along with permutations, solve examples on it and explore their applications in the real world.

For example, suppose we have a team of 15 football. We intend to select a playing team of 11 out of these 15 players. Thus, we want the number of ways in which we can select 11 players out of 15 players. (We are not interested in arranging those 11 players in a row - only the group/combination of those 11 players matters).

## What is Meant by Combinations?

Combinations correspond to the selection of things (and not their arrangement). We do not intend to arrange things. We intend to select them.

Let us make this concept more specific. Suppose we have a set of 6 letters $$\left\{ {A,\,\,B,\,\,C,\,\,D,\,\,E,\,\,F} \right\}$$. In how many ways can we select a group of 3 letters from this set? Suppose we had to find the number of arrangements of 3 letters possible from those 6 letters. That number would be $${}^{\rm{6}}{P_3}$$  . Consider the permutations that contain the letters A, B and C. These are 3! = 6 in number, namely ABC, ACB, BAC, BCA, CAB and CBA.

Now, what we want is the number of combinations and not the number of arrangements. In other words, the 6 permutations listed above would correspond to a single combination. Differently put, the order of things is not important; only the group/combination matters. This means that the total number of combinations of 3 letters from the set of 6 letters available to us would be $${}^{\rm{6}}{P_3}/3!$$  since each combination is counted 3! times in the list of permutations. Thus, if we denote the number of combinations of 6 things taken 3 at a time by $${}^{\rm{6}}{C_3}$$, we have:

${}^{\rm{6}}{C_3} = \frac{{{}^{\rm{6}}{P_3}}}{{3!}}$

In general, suppose we have n things available to us, and we want to find the number of ways in which we can select r things out of these n things.

We first find the number of all the permutations of these n things taken r at a time. That number would be $${}^{\rm{n}}{P_r}$$  .  Now, in this list of $${}^{\rm{n}}{P_r}$$  permutations, each combination will be counted r! times since r things can be permuted amongst themselves in r! ways. Thus, the total number of permutations and combinations of these n things, taken r at a time, denoted by $${}^{\rm{n}}{C_r}$$, will be:

${}^n{C_r} = \frac{{{}^n{P_r}}}{{r!}} = \frac{{n!}}{{r!(n - r)!}}$

Example

Consider the word EDUCATION. This has 9 distinct letters. How many 3-letter permutations (words) can be formed using the letters of this word?

We now know how to answer questions like this; the answer in this particular case will be $$^9{P_3}$$.

Consider the following 3-letter permutations formed using the letters A, E, T from EDUCATION:

AET    ATE

EAT    ETA

TAE    TEA

These 6 different arrangements correspond to the same selection of letters, which is {A, E, T}. Thus, in the list of all 3-letter permutations, we will find that each unique selection corresponds to 6 different arrangements. To find the number of unique 3-letter selections, we divide the number of 3-letter permutations by 6.

Hence, the number of 3-letter selections will be $$\frac{{^9{P_3}}}{6}$$.

## How to Calculate Combinations By Factorials?

Suppose that you have n different objects. You have to determine the number of unique r-selections (selections that contain r objects) which can be made from this group of n objects. Think of a group of n people – you have to find the number of unique sub-groups of size r, which can be created from this group.

The number of permutations of size r will be $$^n{P_r}$$. In the list of $$^n{P_r}$$ permutations, each unique selection will be counted $$r!$$ times, because the objects in an r-selection can be permuted amongst themselves in $$r!$$ ways. Thus, the number of unique selections will be $$\frac{{^n{P_r}}}{{r!}}$$.

$^n{C_r} = \frac{{^n{P_r}}}{{r!}} = \frac{{\left\{ {\frac{{n!}}{{\left( {n - r} \right)!}}} \right\}}}{{r!}} = \frac{{n!}}{{r!\left( {n - r} \right)!}}$

Examples

(i) Out of a group of 5 people, a pair needs to be formed. The number of ways in which this can be done is

$^5{C_2} = \frac{{5!}}{{2!\left( {5 - 2} \right)!}} = \frac{{5!}}{{2!3!}} = \frac{{120}}{{2 \times 6}} = 10$

(ii) The number of 4-letter selections which can be made from the letters of the word DRIVEN is

$^6{C_4} = \frac{{6!}}{{4!\left( {6 - 4} \right)!}} = \frac{{6!}}{{4!2!}} = \frac{{720}}{{24 \times 2}} = 15$

(iii) The number of ways of selecting 2 objects out of n is

\begin{align}&{}^nC_2=\frac{n!}{2!\left(n-2\right)!}=\frac{n\times\left(n-1\right)\times\left(n-2\right)!}{2\left(n-2\right)!}\\&\qquad\qquad\qquad\qquad=\frac{n\left(n-1\right)}2\end{align}

## What Is the General Formula of Combination?

A selection is also called a Combination. We denote the number of unique r-selections or combinations (out of a group of n objects) by $$^n{C_r}$$. Thus,

 $${}^n{C_r} = \frac{{{}^n{P_r}}}{{r!}} = \frac{{n!}}{{r!(n - r)!}}$$

Examples

The number of ways of selecting n objects out of n objects is:

$^n{C_n} = \frac{{n!}}{{n!\left( {n - n} \right)!}} = \frac{{n!}}{{n!0!}} = 1$

The number of ways of selecting 0 objects out of n objects is:

$^n{C_0} = \frac{{n!}}{{0!\left( {n - 0} \right)!}} = \frac{{n!}}{{0!n!}} = 1$

The number of ways of selecting 1 object out of n objects is:

$^n{C_1} = \frac{{n!}}{{1!\left( {n - 1} \right)!}} = \frac{{n \times \left( {n - 1} \right)!}}{{\left( {n - 1} \right)!}} = n$

Important Notes
1. Whenever you read the phrase “number of combinations”, think of the phrase “number of selections”. When you are selecting objects, the order of the objects does not matter. For example, XYZ and XZY are different arrangements, but the same selection.
2. The number of combinations of n distinct objects, taken r at a time (where r is less than n), is $$^n{C_r} = \frac{{^n{P_r}}}{{r!}} = \frac{{n!}}{{r!\left( {n - r} \right)!}}$$.
3. This result above is derived from the fact that in the list of all permutations of size r, each unique selection is counted $$r!$$ times.
4. Out of n objects, the number of ways of selecting 0 or n objects is 1; the number of ways of selecting 1 object is n.
5. Out of n objects, the number of ways of selecting 2 objects is $$^n{C_2} = \frac{{n\left( {n - 1} \right)}}{2}$$.

## Solved Examples

 Example 1

In a party of 10 people, each person shakes hands with every other person. How many handshakes are there?

Solution

Each unique handshake corresponds to a unique pair of persons. Also, note that the order of the two people in the pair does not matter. For example, if X and Y are two people, then XY and YX won’t be counted separately; only the pair {X, Y} will be counted.

Thus, we need to find the number of ways in which 2 people can be selected out of 10. Clearly, the answer is:

$^{10}{C_2} = \frac{{10!}}{{2!\left( {10 - 2} \right)!}} = \frac{{10!}}{{2!8!}} = 45$

 $$\therefore$$ 45 handshakes in total!
 Example 2

How many diagonals are there in a polygon with n sides?

Solution

If we select any two vertices of the polygon and join them, we will get either a diagonal or a side of the polygon. The number of ways of selecting two vertices out of n is $$^n{C_2} = \frac{{n\left( {n - 1} \right)}}{2}$$. Out of these selections, n correspond to the sides of the polygon. Thus, the number of diagonals is:

$\frac{{n\left( {n - 1} \right)}}{2} - n = \frac{{n\left( {n - 3} \right)}}{2}$

 $$\therefore$$ number of diagonals are $$\frac{{n\left( {n - 3} \right)}}{2}$$
 Example 3

A football team has a squad of 15 members. In how many ways can a playing team of 11 players be selected if one particular player must be chosen?

Solution

If we must have one particular player in our team, then we need to choose 10 players out of the remaining 14. This can be done in $$^{14}{C_{10}}$$ ways.

 $$\therefore$$ this can be done in $$^{14}{C_{10}}$$ = 1001 ways
 Example 4

A class has 25 students. For a school event, 10 students need to be chosen from this class. 4 of the students of the class decide that either four of them will participate in the event, or none of them will participate. In how many ways can the selection of 10 students be done?

Solution

Given the specified constraint, we divide the set of all possible selections of 10 students into two groups:

1. The selections include the 4 students; we already have 4 students – we need to select 6 more students out of the remaining 21 students. This can be done in $$^{21}{C_6}$$ ways. Thus, the number of possible selections that include the 4 students is $$^{21}{C_6}$$.
2. The selections do not include the 4 students; now we need to select 10 students out of the remaining 21. This can be done in $$^{21}{C_{10}}$$ ways. Thus, the number of possible selections that do not include the 4 students is $$^{21}{C_{10}}$$.

The total number of selections possible under the specified constraint is $$^{21}{C_6}{ + ^{21}}{C_{10}}$$.

 $$\therefore$$ The answer is $$^{21}{C_6}{ + ^{21}}{C_{10}}$$.
 Example 5

From a basket/plate of 6 apples and 4 mangoes, a total of 5 fruits are needed to be chosen. In how many ways can this be done if the choice has at least 2 apples?

Solution

We divide the situation of “at least 2 apples” into the following cases, and count the number of groups corresponding to each case:

 Case No. of apples No. of mangoes No. of possible selections 1 2 3 $^4{C_2}{ \times ^6}{C_3} = 120$ 2 3 2 $^4{C_3}{ \times ^6}{C_2} = 60$ 3 4 1 $^4{C_4}{ \times ^6}{C_1} = 6$

Thus, the total number of selections consisting of at least 2 apples is 120 + 60 + 6 = 186.

 $$\therefore$$ Total selections are 186

Challenging Questions

Now that you have understood the concepts of permutations and combinations, can you try solving the question that we thought of at the beginning of the article?

Question: What will be the difference in the total number of ways for students in 1 and 2, if:

1. Randomly 4 out of 15 students are made to stand in a circle (or square).
2. Randomly 4 out of 15 students are made to stand in a line.

Hint: Mirror image of the circle will not be a different arrangement of a circle.

## Interactive Questions

Here are a few activities for you to practice.

## Let's Summarize

The mini-lesson targeted the fascinating concept of combinations. The math journey around combination started with what a student already knew and went on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever.

At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!

Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.

Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.

## 1. Does order matter in combinations?

No, the order does not matter in combination. Just the number of selections matters.

## 2. What are the possible combinations of 6 numbers?

For possible combinations (selections) out of 6 different numbers:

Combination of 1 out of 6 is $$^{1}{C_6}$$

Combination of 2 out of 6 is $$^{2}{C_6}$$

Combination of 3 out of 6 is $$^{3}{C_6}$$

Combination of 4 out of 6 is $$^{4}{C_6}$$

Combination of 5 out of 6 is $$^{5}{C_6}$$

Combination of 6 out of 6 is $$^{6}{C_6}$$

## 3. What are the possible combinations out of the digits1234?

The total number of combinations for one digit out of the four is $$^{1}{C_4}$$, while that for two digits is $$^{2}{C_4}$$, while that for three digits is $$^{3}{C_4}$$, and that for four digits is $$^{4}{C_4}$$.

Combinatorics
Combinatorics
Combinatorics
grade 9 | Questions Set 1
Combinatorics
grade 9 | Questions Set 2

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