# Combinations - Definition, Formula, Examples, FAQs (2023)

Combinations are also called selections. Combinations correspond to the selection of things from a given set of things. Here we do not intend to arrange things. We intend to select them. We denote the number of unique r-selections or combinations out of a group of n objects by $$^n{C_r}$$.

Combinations are different from arrangements or permutations. Let us learn more about how to calculate combinations, combinations formula, differences between permutation and combinations, with the help of examples, FAQs.

 1 What Are Combinations? 2 What Is Combinations Formula? 3 Combinations as Selections 4 How to Apply Combinations Formulas? 5 Relationship Between Permutations and Combinations 6 Examples on Combinations 7 FAQs on Combinations

## What Are Combinations?

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 $$^n{C_r}$$ and it is given by, $$^n{C_r} = \dfrac{n!}{r!(n-r)!}$$,where 0 ≤r ≤n.This forms the general combination formula which is$$^n C_r$$ formula.

This formula to find the number of combinations by using r objects from the n objects, is also referred as the ncr formula.

## What Is Combinations Formula?

The combinations formula is used to easily find the number of possible different groups of r objects each, which can be formed from the available n different objects. Combinations formula is the factorial of n, divided by the product of the factorial of r, and the factorial of the difference of n and r respectively.

### Combinations Formula: $$^nC_r = \dfrac{n!}{r!.(n - r)!}$$

The combinations formula is also referred to as ncr formula. To use the combinations formula we need to know the meaning of factorial, and we have n! = 1× 2× 3× .... (n - 1)× n.

## Combinations as Selections

Suppose we have a set of 6 letters { A,B,C,D,E,F}. In how many ways can we select a group of 3 letters from this set? Suppose wefind the number of arrangements of 3 letters possible from those 6 letters. That number would be 6P$$_3$$. Consider the permutations that contain the letters A, B, and C. These are 3! = 6 ways, 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 now in our selection.This means that the total number of combinations of 3 letters from the set of 6 letters available to us would be 6P$$_3$$/3! ways,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 6C$$_3$$, we have:

$$^6{C_3}= \dfrac{^6{P_3}}{3}$$. This is also said as 6 choose 3.

A few important results on combinations are as follows:

• 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: $$^n{C_1} = \frac{{n!}}{{1!\left( {n - 1} \right)!}} = \frac{{n \times \left( {n - 1} \right)!}}{{\left( {n - 1} \right)!}} = n$$
• $$^n{C_r} = ^n{C_{n-r}}$$
• $$^n{C_r} +^n{C_{r-1}} = ^{n+1}{C_{r}}$$

## How To ApplyCombinations Formula?

We calculate combinations using the combinations formula, and by usingfactorials and in terms of permutations. 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 $$^nP_r$$ . Now, in this list of $$^nP_r$$ 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 $$^nC_r$$, will be:

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

(Video) Combinations Definition and Formula

## Relationship Between Permutations and Combinations

Permutation and combination formulas and concepts have a lot of similarities. 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 combinations canbe $$\frac{{^n{P_r}}}{{r!}}$$.

$$^n{C_r} = \dfrac{^nP_r}{r!} = \dfrac{n!}{(n - r) r!} = \dfrac{n!}{r!(n - r)}$$

## Examples on Combinations

Example 1: Consider the word EDUCATION. Thishas 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 Combinationcorresponds to 6 different arrangements. To find the number of unique 3-letter selections, we divide the number of 3-letter permutations by 6.

(Video) Combination formula-Examples and How to Solve

Hence, the number of 3-letter selections will be $$\dfrac{^9P_3}{6}$$ = 60480/ 6 = 10,080

Example 2: Out of a group of 5 people, a pair needs to be formed. The number of possible combinations can be calculated as follows.

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

Example 3:The number of 4-letter Combinationswhich can be made from the letters of the word DRIVEN is

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

Important Notes

• 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 have the same selection.
• The number of combinations of n distinct objects, taken r at a time (where r is less than n), is $$^nC_r = \dfrac{^nP_r}{r}$$ = $$\dfrac{n!}{r! (n-r)!}$$
• 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.
• Out of n objects, the number of ways of combinations0 or n objects is 1; and the number of ways of selecting 1 object or (n - 1) object is n.
• Out of n objects, the number of ways of selecting 2 objects is $$^n{C_2} = \frac{{n\left( {n - 1} \right)}}{2}$$.

Also Check:

• Fundamental Counting Principle
• Permutations as Arrangements
• Permutations and Combinations formula

## FAQs on Combinations

### What Are Combinations In Numbers?

Combinations are selections. Selecting r objects out of the given n objects is given by using the factorials. It is denoted by $$^nC_r = \dfrac{n!}{r!.(n - r)!}$$. The combinations are the different subgroups that can be formed from the given larger group of objects.

### How To Use TheCombinations Formula?

Combinations are calculated using the combination formula $$^nC_r = \dfrac{n!}{r! (n-r)!}$$, while we need to choose r items out of n items. Here we use the factorial formula of n! = 1× 2× 3×.......(n - 1)× n.

### Does order matter in combinations?

No, the order does not matter in combination. Just the number of selections or subgroupsmatters. The number of dresses in the wardrobe can be selected at random in order. Picking 2 clothes out of 8 from the wardrobe requires $$^8C_2$$ ways = $$\dfrac{8!}{2! 6!}$$ = 28 ways. Here we consider the set of two and do not look into the order of the selection.

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

The possible combinations (selections) out of 6 different numbers are as follows:

• Combination of 1 out of 6 is $$^{6}{C_1}$$
• Combination of 2 out of 6 is $$^{6}{C_2}$$
• Combination of 3 out of 6 is $$^{6}{C_3}$$
• Combination of 4 out of 6 is $$^{6}{C_4}$$
• Combination of 5 out of 6 is $$^{6}{C_5}$$
• Combination of 6 out of 6 is $$^{6}{C_6}$$

### What are the possible combinations out of the digits 1234?

The total number of combinations are possible this way: choosing 1 digit out of 4, choosing 2 digits out of 4, choosing 3 digits out of four and choosing 4 digits out of 4 = $$^{4}{C_1}+ ^{4}{C_2} + ^{4}{C_3} + ^{4}{C_4}$$ = 4 + 6 + 4 + 4 = 18 ways

### How Are Permutations and Combinations Related?

The permutations and combinations are relatedusing the combination formula $$^n{C_r} = \dfrac{^nP_r}{r!} = \dfrac{n!}{(n - r) r!} = \dfrac{n!}{r!(n - r)}$$. The combinations are the selection of r things taken from n different things, and permutation is the different arrangement of those r things.

### What Are The Differences Between Permutations and Combinations?

Permutations are seen as arrangements of r things out of n things, whereas combinations are seen as selections of r things out of n things. $$^nP_r =\dfrac{n!}{r!}$$ and $$^nC_r = \dfrac{n!}{r! (n-r)!}$$. For the given r things out of n things, the number of permutations are greater than the number of combinations.

(Video) Permutations and Combinations Tutorial

### What Are the Combination Examples?

The combination examples include the groups formed from dissimilar obects.The formation of a committee, the sport team, set of different stationary objects, team of people are some of the combination examples.

(Video) Combination formula | Probability and combinatorics | Probability and Statistics | Khan Academy

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