Combinations: Definition, Formula, Solved Examples

Combinations: You may have noticed that people around you have specific combinations as their preferences. For example, my father likes to eat dosa with sambar, and my sister likes to wear white shoes with her blue denim. While mom always makes phulkas with potatoes, I like them better with dal.

Likewise, there are multitudes of combinations around us. But how can we pick over the other? How does this work in Mathematics? This article will answer these questions. It will introduce, define, and explain the concept of combinations. Let’s also learn to compare combinations with permutations.

What are Combinations?

Mathematically, a combination is a process of selecting objects from the given options.

Example: Imagine there is a breakfast buffet. The dishes available are listed below in set B. If you can select any \(3\) dishes out of the \(5\) available, in how many different ways can you fill your plate?
\(B = \left\{ {{\text{Dosa,}}\,{\text{Idly,}}\,{\text{Sambar,}}\,{\text{Chutney,}}\,{\text{kesari}}} \right\}\)

The different ways of filling your plate with any \(3\) dishes are shown below:

Therefore, there are \(10\) different ways to select \(3\) out of \(5\) dishes.

Here, each selection is called a combination. Hence, there are \(10\) possible combinations.

Define Combinations

Combinations can be defined as selecting some or all of the objects from a given collection without paying attention to the order. It is denoted as \(^n{C_r}\), where \(n\) is the number of given objects, and \(r\) is the number of objects to be selected. Combinations is also known as selection.

While combinations are all about grouping, permutations are about arranging objects from a collection. We will further discuss the differences and relations between these two counting concepts.

Number of Combinations

As shown in the breakfast example, it is possible to manually count the number of combinations for smaller sets of objects. As the size of the set increases, the possibility of error increases when done manually. This is when the formula comes into play. The number of combinations is the ratio of the factorial of the total number of objects to the product of the factorial of the required number of objects and the factorial of their difference.

General Formula of Combinations

To select \(r\) objects from the given \(n\) objects, the number of combinations is given by
\(^n{C_r} = \frac{{n!}}{{r!\left( {n – r} \right)!}}\)
Where, \(r\) is a non-negative number that is less than or equal to \(n\).
This is also called the \(^n{C_r}\) formula.

Illustration

You visit a neighbourhood café with your \(4\) friends. There are \(10\) items on their menu. If you each order one item, in how many different ways can you place the order? The menu of the café is given below.

Total number of items on the menu, \(n=10\)
Items to be selected, \(r=5\)
Number of combinations, \(^{10}{C_5} = \frac{{10!}}{{5!\left( {10 – 5} \right)!}}\)
\( = \frac{{10!}}{{5!5!}}\)
\( = \frac{{6 \times 7 \times 8 \times 9 \times 10}}{{5!}}\)
\( = \frac{{6 \times 7 \times 8 \times 9 \times 10}}{{1 \times 2 \times 3 \times 4 \times 5}}\)
\( = \frac{{7 \times 2 \times 9 \times 10}}{5}\)
\(\therefore {\,^{10}}{C_5} = 252\)
So, there are \(252\) different ways in which you and your friends can place an order of \(5\) items from the menu.

Representation of Combinations

We know that combinations is represented as \(^n{C_r}\). The other common representations of \(^n{C_r}\) are:

• \(C\left( {n,\,r} \right)\)
• \(_r^nC\)
• \(_n{C_r}\)
• \(\left( \begin{gathered}  n \hfill \\  r \hfill \\\end{gathered}  \right)\)

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

Permutations vs Combinations

Have you seen a combination lock? It usually requires a specific \(4-\) digit number to open when locked. But here, although we know that we need only \(4\) numbers, the order of numbers is extremely important. For example, if the number to unlock is \(4532\), entering \(2345\) or \(5432\) will not open the lock. Such cases where the order is essential is called permutations. Ideally, a combination lock should be called a permutation lock.

The table below shows the comparison of permutations and combinations.

Relation between Permutations and Combinations

We know that the permutations is given by \(^n{P_r} = \frac{{n!}}{{\left( {n – r} \right)!}}\), and combinations is given by the formula \(^n{C_r} = \frac{{n!}}{{r!\left( {n – r} \right)!}}\), where \(r\) is the number of objects taken from a collection of \(n\) objects.
Formula for Combinations, \(^n{C_r} = \frac{{n!}}{{r!\left( {n – r} \right)!}}\)
\( \Rightarrow r!\, \times {\,^n}{C_r} = \frac{{n!}}{{\left( {n – r} \right)!}}\)
\(\therefore {\,^n}{P_r} = r!\, \times {\,^n}{C_r}\,\)
\( \Rightarrow \frac{{^n{P_r}}}{{r!}}{ = ^n}{C_r},\,0 \leqslant r \leqslant n\)
Hence, we can say that permutations is always greater than combinations for the same values of \(r\) and \(n\).

Take-aways

1. When \(r=n\), the number of combinations is given by,
   \(^n{C_n} = \frac{{n!}}{{n!\left( {n – n} \right)!}}\)
   \( = \frac{{n!}}{{n!0!}}\)
   \(\therefore {\,^n}{C_n} = 1\)
2. The number of combinations to select \(0\) objects from a collection of \(n\) objects is \(1\). This is because combinations is the different number of ways of selecting some or all the objects in the collection. Selection of nothing means that we are leaving behind all the objects and there is only one way of doing it.
   When \(r=0\), the number of combinations is given by,
   \(^n{C_0} = \frac{{n!}}{{0!\left( {n – 0} \right)!}}\)
   \( = \frac{{n!}}{{n!}}\)
   \(\therefore {\,^n}{C_0} = 1\)
   This means that when zero objects are taken from a collection of \(n\) objects, the number of combinations is \(1\).
3. Selecting \(r\) objects from a collection of \(n\) objects is the same as selecting \(n-r\) objects.
   Proof: Let \(r=n-r\). Then,
   \(^n{C_{n – r}} = \,\,\frac{{n!}}{{\left( {n – r} \right)!\left( {n – \left( {n + r} \right)} \right)!}}\)
   \( = \,\,\frac{{n!}}{{\left( {n – r} \right)!\left( {n – n + r} \right)!}}\)
   \( = \,\,\frac{{n!}}{{\left( {n – r} \right)!r!}}\)
   \(\therefore {\,^n}{C_{n – r}}{ = ^n}{C_r}\)
   Hence, proved.
4. If \(^n{C_a} = {\,^n}{C_b}\). Then,
   \(a=b\)
   Since \(^n{C_b} = {\,^n}{C_{n – b}}\), we can say that \(b=n-b\).
   \(\therefore \,a = n – b\)
   \( \Rightarrow n = a + b\)
5. \(^n{C_r} + {\,^n}{C_{r – 1}}{ = ^{n + 1}}{C_r}\)
   Proof:
   L.H.S. \(^n{C_r} + {\,^n}{C_{r – 1}} = \frac{{1}}{{r!\left( {n – r} \right)!}} + \frac{{1}}{{\left( {r – 1} \right)!\left( {n – \left( {r – 1} \right)} \right)!}}\)
   \( = n!\left( {\frac{{n!}}{{r!\left( {n – r} \right)!}} + \frac{{n!}}{{\left( {r – 1} \right)!\left( {n – r + 1} \right)!}}} \right)\)
   \( = n!\left( {\frac{1}{{\left( {r – 1} \right)! \times r \times \left( {n – r} \right)!}} + \frac{1}{{\left( {r – 1} \right)!\left( {n – r + 1} \right)!}}} \right)\)
   \( = \frac{{n!}}{{\left( {r – 1} \right)!}}\left( {\frac{1}{{r\left( {n – r} \right)!}} + \frac{1}{{\left( {n – r + 1} \right)\left( {n – r} \right)!}}} \right)\)
   \( = \frac{{n!}}{{\left( {r – 1} \right)!\left( {n – r} \right)!}}\left( {\frac{1}{r} + \frac{1}{{\left( {n – r + 1} \right)}}} \right)\)
   \( = \frac{{n!}}{{\left( {r – 1} \right)!\left( {n – r} \right)!}}\left( {\frac{{n – r + 1 + r}}{{r\left( {n – r + 1} \right)}}} \right)\)
   \( = \frac{{n!\left( {n + 1} \right)}}{{r\left( {r – 1} \right)!\left( {n – r} \right)!\left( {n – r + 1} \right)}}\)
   \( = \frac{{\left( {n + 1} \right)!}}{{r!\left( {n – r + 1} \right)!}}\)
   \({ = ^{n + 1}}{C_r}:\) R.H.S
   Hence, proved.

Solved Examples

Q1. In how many ways can you pick \(3\) clothes out of \(6\)?
Solution:
To select \(r\) objects from a set of \(n\),\(^n{C_r} = \frac{{n!}}{{r!\left( {n – r} \right)!}}\)
Here, \(n=6\) and \(r=3\)
\(\therefore {\,^6}{C_3} = \frac{{6!}}{{3!\left( {6 – 3} \right)!}}\)
\( = \frac{{6!}}{{3! \times 3!}}\)
\( = \frac{{6 \times 5 \times 4}}{{3 \times 2 \times 1}}\)
\( = 2 \times 5 \times 2\)
\(=20\)

Q2. How many \(9-\) letter combinations can be made using the letters of the word ‘UNCOPYRIGHTABLE’?
Solution:
Number of letters in the given word \(=15\)
Number of letters in the required combinations \(=9\)
Number of combinations of \(9-\) letter words \({ = ^{15}}{C_9}\)
\(^{15}{C_9} = \frac{{15!}}{{9!\left( {15 – 9} \right)!}}\)
\( = \frac{{15!}}{{9!\, \times 6!}}\)
\( = \frac{{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7}}{{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}\)
\( = \frac{{15 \times 14 \times 13 \times 12 \times 11 \times 10}}{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}\)
\( = \frac{{5 \times 7 \times 13 \times 3 \times 11 \times 2}}{{6 \times 1}}\)
\( = \frac{{30030}}{6}\)
\(=5005\)

Q3. There were \(66\) handshakes at a gathering. It was confirmed that everybody shook hands with everybody else. How many people were there at the gathering?
Solution:
Total number of handshakes, \(^n{C_r} = 66\)
Number of hands in a shake, \(r=2\)
\(\therefore \) Total number of people can be calculated using
\(^n{C_2} = \frac{{n!}}{{2!\left( {n – 2} \right)!}} = 66\)
\( \Rightarrow \frac{{n\left( {n – 1} \right)}}{2} = 66\)
\({n^2} – n = 132\)
\( \Rightarrow {n^2} – n – 132 = 0\)
\({n^2} – 12n + 11n – 132 = 0\)
\(n\left( {n – 12} \right) + 11\left( {n – 12} \right) = 0\)
\(\left( {n + 11} \right)\left( {n – 12} \right) = 0\)
Since, the number of people cannot be negative, \(n=12\)

Q4. If there are \(6\) men and \(5\) women, how many different subgroups of \(5\) people can be formed?
Solution:
Number of men \(=6\)
Number of women \(=5\)
Total number of people, \(n=11\)
Number of people in the subgroup, \(r=5\)
\(\therefore \) Possible number of subgroup combinations \({ = ^{11}}{C_2}\)
\( = \frac{{11!}}{{5!\left( {11 – 5} \right)!}}\)
\( = \frac{{11!}}{{5!\, \times 6!}}\)
\( = \frac{{7 \times 8 \times 9 \times 10 \times 11}}{{1 \times 2 \times 3 \times 4 \times 5}}\)
\( = \frac{{7 \times 2 \times 3 \times 2 \times 11}}{2}\)
\(=462\)

Q5. What is the number of ways of choosing \(4\) cards from a pack of \(52\) cards? In how many of these
a. \(4\) cards of the same suit?
b. \(4\) cards of different suits?
c. are face cards?
d. \(2\) are red, and \(2\) are black cards?
e. cards of one colour?
Solution:
The number of ways to select \(4\) cards from a pack of \(52\) is given by,
\(^n{C_r} = \frac{{n!}}{{r!\left( {n – r} \right)!}}\)
Here, we know that \(r=4\) and \(n=52\).
\({ \Rightarrow ^{52}}{C_4} = \frac{{52!}}{{4!\left( {52 – 4} \right)!}}\)
\( = \frac{{52!}}{{4!\, \times 48!}}\)
\( = \frac{{52 \times 51 \times 50 \times 49}}{{4!\,}}\)
\( = \frac{{13 \times 17 \times 25 \times 49}}{{1\,}}\)
\(\therefore {\,^{52}}{C_4} = 270725\)
a. Choose \(4\) cards of the same unit
Number of suits \(=4\)
Number of cards in each suit \(=13\)
Number of ways to select \(4\) cards of the same suit \({ = ^{13}}{C_4}\)
\(\therefore \) Total number of ways to select \(4\) cards from the pack \({ = ^{13}}{C_4}{ + ^{13}}{C_4}{ + ^{13}}{C_4}{ + ^{13}}{C_4}\)
\( = 4{ \times ^{13}}{C_4}\)
\( = 4 \times \,\frac{{13!}}{{4!\left( {13 – 4} \right)!}}\)
\( = 4 \times \,\frac{{13!}}{{4!\, \times 9!}}\)
\( = 4 \times \,\frac{{10 \times 11 \times 12 \times 13}}{{1 \times 2 \times 3 \times 4}}\)
\( = \,\frac{{10 \times 11 \times 12 \times 13}}{{2 \times 3}}\)
\(=2860\)
b. Choose \(4\) cards of different suits
Number of suits \(=4\)
Number of cards in each suit \(=13\)
Number of ways to select \(1\) cards of the same suit \({ = ^{13}}{C_1}\)
\(\therefore \)Total number of ways to select  \(1\) card from each suit\({ = ^{13}}{C_1}{ \times ^{13}}{C_1}{ \times ^{13}}{C_1}{ \times ^{13}}{C_1}\)
\( = {\left( {^{13}{C_1}} \right)^4}\)
\( = {13^4}\)
\(=28561\)
c. Choose \(4\) face cards
Number of face cards \(=12\)
\(\therefore \) Total number of ways to select \(4\) face cards from a set of \(12\)\({ = ^{12}}{C_4}\)
\( = \frac{{12!}}{{4!\left( {12 – 4} \right)!}}\)
\( = \frac{{12!}}{{4!\, \times 8!}}\)
\( = \,\frac{{9 \times 10 \times 11 \times 12}}{{1 \times 2 \times 3 \times 4}}\)
\( = \,\frac{{3 \times 5 \times 11 \times 3}}{1}\)
\(=495\)
d. Choose \(2\) red cards and \(2\) black cards
Number of red cards \(=26\)
Number of ways to select \(2\) red cards \({ = ^{26}}{C_2}\)
Number of black cards \(=26\)
Number of ways to select \(2\) black cards \({ = ^{26}}{C_2}\)
\(\therefore \) Total number of ways to select \(2\) red and \(2\) black cards \({ = ^{26}}{C_2}{ \times ^{26}}{C_2}\)
\( = {\left( {^{26}{C_2}} \right)^2}\)
\( = {\left( {\frac{{26!}}{{2!\left( {26 – 2} \right)!}}} \right)^2}\)
\( = {\left( {\frac{{26!}}{{2!\, \times 24!}}} \right)^2}\)
\( = {\left( {\frac{{25 \times 26}}{{1\, \times 2}}} \right)^2}\)
\( = {\left( {25\, \times 13} \right)^2}\)
\( = {325^2}\)
\(=105625\)
e. Choose \(4\) cards of one colour
Number of red cards \(=26\)
Number of ways to select \(4\) red cards \({ = ^{26}}{C_4}\)
Number of black cards \(=26\)
Number of ways to select \(4\) black cards \({ = ^{26}}{C_4}\)
\(\therefore \) Total number of ways to select \(4\) cards of the same colour \({ = ^{26}}{C_4}{ + ^{26}}{C_4}\)
\( = 2\,{ \times ^{26}}{C_4}\)
\( = 2\, \times \frac{{26!}}{{4!\left( {26 – 4} \right)!}}\)
\( = 2\, \times \frac{{26!}}{{4!\, \times \,22!}}\)
\( = 2\, \times \frac{{23 \times 24 \times 25 \times 26}}{{4\, \times 3 \times \,2 \times 1}}\)
\( = 2\, \times \frac{{23 \times 2 \times 25 \times 13}}{1}\)
\(=29900\)

Summary

A combination is a mathematical concept to determine the number of possible arrangements of \(n\) objects taken \(r\) at a time. It is represented as \(^n{C_r}\). It is different from permutations because the order of objects in combinations does not matter. The number of combinations is the ratio of the factorial of the total number of objects to the product of the factorial of the required number of objects and the factorial of their difference. The number of permutations is \(r!\) times the number of combinations. Hence, the number of combinations is always lesser than the number of permutations for the same values of \(r\) and \(n\).

Frequently Asked Questions (FAQs)

Q1. How do you use combinations?
Ans: Combinations is used to calculate the possible number of selections of \(n\) objects taken \(r\) at a time. It is denoted by \(^n{C_r}\).

Q2. What do you mean by permutations?
Ans: Permutations is the possible number of arrangements of \(n\) objects taken \(r\) at a time. It is denoted by \(^n{P_r}\) .

Q3. How do you add two combinations?
Ans: To add two combinations, the value of the individual combinations is added.

Q4. How many combinations can you make with 3 numbers?
Ans: Here, \(n=3\), \(r=3\) and \(n=r\). Therefore, \(^n{C_n} = 1\).

Q5. How do you calculate combinations?
Ans: Combinations of \(r\) objects from a collection of \(n\) objects is calculated using the formula:
\(^n{C_r} = \frac{{n!}}{{r!\left( {n – r} \right)!}}\)