The Combination Formula is a selection procedure used in mathematics to identify the most probable arrangements among a bunch of elements. However, the sequence in which the items are chosen is irrelevant in this strategy. You can, however, choose as many items as you wish, in any order. However, this is why the Combination formula is also known as the Selection formula. When games of chance began to appear in the early 17th century, the combination formula and probability were introduced. Blaise Pascal and Pierre de Fermat, two French mathematicians, were the first to present this mathematical technique.

Table of Contents

**Combination Formula in maths**

When you look at the equation for calculating combinations, you’ll notice that it employees factorials all over the place. Remember, to calculate combinations, use the formula nCr = n! / r!(n – r)!, where n denotes the total number of things and r is the number of items we choose at a given moment. Let’s have a look at an example of how to do a combination calculation.

This week, 10 new DVD releases are available for rental. However, this weekend, John wants to watch three movies. How many movie combinations can he choose from?

In this challenge, John must select three films from a list of ten new releases. However, we represent the variable n by 10 and the r variable by 3. As a result, our equation would be 10C3 = 10! / 3!(10 – 3)! .

The first step is to deduct 10 from the bottom of this equation. Because 10 – 3 = 7, our equation is 10! / 3!7!

Following that, we must expand each of our factorials. On the top, 10! would equal 10×9×8×7×6×5×4×3×2×1, and 3!7! would equal 3×2×1×7×6×5×4×3×2×1. However, the simplest solution to this problem is to eliminate like phrases. On both the top and bottom of our equation, we can see a 7, 6, 5, 4, 3, 2 and 1.

We can remove these conditions. However, we can now see that we have 10×9×8 left on top and 3×2×1 left on the bottom of our equation. However, we can simply multiply from here. 10×9×8 equals 720, and 3×2×1 equals 6. As a result, our equation is now 720/6.

We’ll divide 720 by 6 to get 120 as the answer to this problem. John now knows that he has the option of choosing from 120 distinct new-release movie combinations this week.

**Combination Formula nCr**

nCr = n! / r!(nr)! is the formula for solving combination problems. However we can use this formula with any combination that does not allow for repetition.

Where n is the number of things to choose from and r is the number of things we choose from n of them, there is no repetition and order is irrelevant.

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Simply enter the number of values in the formula and multiply by factorials to get the result. By multiplying all whole numbers from our selected number down to one, we can solve the factorial. Reduce the answer by solving the factorial.

**Combination Formula Calculator**

Many resources exist on the internet that can assist you in calculating the number of possible combinations from a group of objects. A simple combination calculator is the Online Combination Formula Calculator. Users only need to enter the numbers n and r in the forms, and the combination will be calculated. However, this saves time and eliminates the need for manual calculations.

- Combination calculator websites can also assist in determining a combination.
- These websites are simple to navigate.
- It saves time and simplifies calculations.

**Combination Formula in Excel**

MS Excel can also be used to calculate the combination. However, the COMBIN function in Microsoft Excel yields the number of combinations for a given number of items. However, the following is the COMBIN formula: COMBINATION ( number, chosen )

Where number denotes the total number of things, and chosen denotes the number of items in a given combination that have been chosen. A numeric number will always be returned by the COMBIN function.

- MS Excel also has a capability for solving combination problems.
- The COMBIN function is employed.
- The solution is a numerical value.
- It’s easy to use.

**Combination Formula Definition**

A combination formula is a method for calculating the number of alternative arrangements given a set of items or numbers.

The combination formula, on the other hand, assists us in selecting a required item from a set of options. However, we can create a subset using the combination approach without having to replace any numbers or items.

The number of alternative combinations of “r” objects from “n” separate objects can be expressed as follows:

nCr = C(n,r) = n! / r! × (n – r)!

Where,

- 0 ≤ r ≤ n.
- n denotes the total number of elements present in a set.
- k denotes the number of selected objects from the set.
- ! Stands for factorial

All positive integers smaller or equal to the number in front of the factorial sign are represented by the factorial (noted as “!”). 4! = 1 x 2 x 3 x 4 = 24 is an example.

Note that the formula above may only be utilized when selecting objects from a collection without repetition.

The combination formula, on the other hand, determines the combination of r objects picked at a time from n without repetition.

If you have four types of paint in your hand (n = 4) and can only choose three (r = 3), the number of ways to choose the paints with repetition can be estimated as 4C3.

**Combination Formula Properties**

- nCr is a natural number.
- nC0 = (nCn) = 1
- nC1 = n
- nCr = nCn-r
- nCx = nCy means, x = y or, x+y = n
- n (n-1Cr-1) = (n – r + 1) × nCr-1
- nCr = (n/r)×(n-1Cr-1)
- (nCr / nCr-1) = (n-r+1) / r
- nC0 + nC1 + nC3 + …… + nCn = 2^n
- nC0 + nC2 +…. = nC1 + nC2 +…. = 2^(n-1)
- (2n+1)C0 + (2n+1)C1 + … + (2n+1)Cn = 2^2n
- on + (n-1)Cn + ….. + (2n-1)Cn = 2nCn+1

**Combination Formula Derivation**

Multiplication and permutation can thus be used to construct the combination formula: n methods to select the first thing from n different objects

Number of options for selecting a second object from a set of (n-1) different objects: (n-1) options

Number of options for selecting a third object from a set of (n-2) distinct objects: (n-2) options

There are a total of (n-3) different ways to choose a fourth object: (n-3) ways to choose rth object from (n-(r-1)) distinct objects: (n-(r-1)) ways to choose r^{th} object from (n-(r-1)) distinct objects: (n-(r-1)) ways to choose r^{th} object from (n-(r-1)) distinct objects: (n

An ordered subset of r elements is created by selecting r things from the initial set of n things.

∴ the number of ways to make a selection of r elements of the original set of n elements is [n (n – 1),(n – 2),(n-3),(n – (r – 1))] or [n (n – 1),(n – 2),(n – r + 1) ]

Let’s take a look at the ordered subset of r elements, as well as all of its variations. However, this sub-total set’s number of permutations is equal to r! Because every combination of r objects can be rearranged in r! different ways.

As a result, nCr r is the total number of permutations of n different objects taken r at a time. However, it is, on the other hand, nPr.

nPr=nCr×r!

**Combination Formula in Probability **

In probability theory and other branches of mathematics, a combination is a set of outcomes whose order is irrelevant. When ordering a pizza, for example, it makes no difference whether you order it with ham, mushrooms, and olives or olives, mushrooms, and ham. Meanwhile, you’re going to get the same pizza as me!

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When calculating probabilities, it’s common to need to figure out how many possible combinations there are.

I’ll show you how to calculate the number of combinations with and without repetition in this post, as well as teach you how to write combinations in standard notation. I’ll also walk you through some practise issues so you can understand how the process works. Meanwhile, you’ll discover how to calculate your chances of winning the Powerball lottery jackpot!

**Comparing Combinations to Permutations**

Combinations and permutations are two different notions in probability theory.

Combinations: The sequence of the outcomes is irrelevant.

Permutations: The sequence in which events occur is important.

To demonstrate both notions, let’s use three letters: C, J, and M.

Assume we’re utilizing these letters to represent team members such as Carl, Jack, and Mary. C, J, M is the same as M, J, C when it comes to team members. Then, in any case, you’ll be working with the same people. As a result, because you’re on the same side, the order of the letters doesn’t matter. However, these three letters make up a single word.

If you’re using these letters for a short passcode, though, the order of the letters matters. MJC will not work if the right code is CJM.

One conceivable combination of these three letters is CJM. Meanwhile, you can make 33 = 27 different permutations with just these three letters.

The number of permutations exceeds the number of combinations when there are at least two of them.

I’m simply working with combos in this post. Read my post on Using Permutations to Calculate Probabilities to learn more about permutations.

**Combination Formula with Repetition**

Assume that we have an n-element set A. A combination of n objects taken r at a time with repetition is any selection of r objects from A in which each object can be selected several times. aaa, aab, aac, aad, abb, abc, abd, acc, acd, add, dbb, bbc, bbd, bcc, bcd, bdd, ccc, ccd, cdd, ddd, ccc, ccd, ddd, ccc, ccd Now, if two combinations with repetition contain the same elements repeated the same number of times, regardless of their sequence, we consider them to be identical.

It’s worth noting that the following are interchangeable:

- The total number of n item combinations taken r at a time with repetition.
- The number of possible distributions of r identical objects among n different containers.
- Now consider how many nonnegative integer solutions there are to the equation:

x_{1} + x_{2} + · · · +in_{n} = r.

The combination formula, on the other hand, does not allow for repetition.

Ex.1 A particular password is made up of three separate letters of the alphabet, each of which is used only once. How many distinct password combinations are there?

Ans.

In a password, the order is important, and the problem states that you cannot repeat letters. As a result, permutations without repeated formulas are required.

As a result, there are the following permutations of three letters picked from a list of 26.

n! / (n-r)! = 26! / (26-3)! = 15,600 passwords.

**Combination formula without Repetition**

You’re working with combinations without repetition when the results don’t repeat. As a starting point, these calculations employ the following equation for permutations without repetition.

n!/ (n-r)!

The order matters in permutations. Meanwhile, combinations without repetition, do not care about the order and produce a lower number than permutations. However, the first part of the equation captures the number of permutations in which the order matters. Then, the second part lowers that number since the order of the elements in a combination is irrelevant. However, the final equation is the product of the two.

n! / (n-r)! × 1/r! = n! / r! ( n – r )!

In the simplified form:

nCr = n! / r! ( n – r )!

Where:

n = the number of possible outcomes at the start.

r = the size of each combination.

Consider a method that selects five people from a pool of 30 to form a team. However, it makes no difference in which order the method selects team members. Assume A, B, C, D, and E make up a team. That’s the same as putting together a team with the same members but in a different sequence. However, there is also no recurrence because the method can only select one person at a time.

Because we’re selecting from a pool of 30 individuals, n = 30 in this case. Because the squad can only have five members, R = 5. Adding those figures to the equation:

Example of a calculation for non-repetitive pairings.

30C5 = 142,506 using standard notation

When you draw from a pool of 30 people, you get 142,506 different five-person teams.

You may also use Pascal’s triangle to get the number of possible combinations without having to repeat the process.

**Using Combinations to Calculate Probabilities**

We estimated the number of options for various scenarios in the preceding examples. However, we didn’t utilize them to calculate probabilities, though. A probability is defined as the number of combinations judged to constitute an occurrence divided by the total number of combinations in this context.

If you’re given a probability problem involving combinations, you’ll need to figure out a few things before you can solve it.

Calculate the probability using a ratio.

Determine if the numerator and denominator require combinations, permutations, or a mix of the three. However, we’ll stick to combos for this post.

Are these repeating or non-repeating combinations?

You must identify the n and r to enter into the equations for both types of combinations.

Let’s have a look at an example of a probability calculation that uses combinations.

Problem: In the Powerball lottery, what is the likelihood of winning both the basic game and the jackpot?

**Probability of winning the basic game in the Powerball lottery**

A lottery is a famous example of probability employing combinations without repetition, in which machines select balls with numbers at random from a pool of balls. However, it makes no difference what order the numbers are in. All you have to do now is match the digits. However, there is no repetition because we can draw each ball only once.

We’ll use the Powerball lottery as an example, which is available in 45 states across the United States. However, there are 69 white balls in the basic game, and a machine selects five of them at random. However, we define the likelihood of victory by the following ratio:

**Probability = Winning combination / Total number of combinations**

The numerator is straightforward. However, there is just one winning combination that does not need to be repeated. Meanwhile, you can compute 5C5 using the equation, which equals the number of winning combinations with five numbers in the winning combination and five winning numbers to choose from. However, it makes obvious sense. Only one winning combination exists.

Calculate 69C5 for the denominator, which is the number of combinations when you draw five numbers from a total of 69 numbers. Let’s plug these figures into the formula:

69C5 = 11,238,513

There are 11,238,513 possible combinations when you draw five numbers from a list of 69.

The following is the customary notation for the probability of winning the basic game in Powerball:

5C5 / 69C5 = 1/11,238,513 = 0.000000089

You have a 1 in 11,238,513 chance of winning the basic game in the Powerball lottery.

**The probability of winning the jackpot in Powerball**

To win the Powerball jackpot, you must first win the basic game and then match one red Powerball number. However, the Powerballs are 26 red balls that we draw at random from a separate machine. Let’s see how likely it is that you’ll win the lottery!

For the Powerball’s numerator, there is only one winning combination. For 26C1 = 26, we must compute the number of combinations in the denominator. However, this is unsurprising. When you choose one number from 26 possibilities, there are 26 possible combinations! As a result, the following is the probability of matching the Powerball:

1C1/ 26C1 = 1/26 = 0.038

To account for this in the basic game’s probability calculations, we must multiply the two ratios using the multiplication method for independent occurrences.

Using conventional notation for combinations, the odds of winning the Powerball jackpot are as follows:

(5C5) (1C1) / (69C5) (26C1) = 1/292,201,338 = 0.0000000034

You have a 1 in 292,201,338 chance of winning the jackpot in the Powerball lottery

**Combination Formula with example **

## Example 1

**Mr. Smith is the chair of a committee. How many ways can a committee of 4 be chosen from 9 people given that Mr. Smith must be one of the people selected?**

**Ans.**

We’ve already chosen Mr. Smith, so we’ll have to pick three more persons from a pool of eight. Because the order does not matter while selecting a committee, we must use the combination without repetition formula.

Hence, according to the combination formula,

n! / {r! (n-r)!} = 8! / 3!(8-3)! = 56 ways.

## Example 2

**A student has to answer 10 questions, choosing at least 4 from each of Parts A and B. If there are 6 questions in Part A and 7 in Part B, in how many ways can the student choose 10 questions?**

Ans. However, the possibilities are:

4 from Part A and 6 from Part B

or 5 from Part A and 5 from Part B

or 6 from Part A and 4 from Part B.

Therefore, according to the combination formula, the required number of ways is

6C4 × 7C6 + 6C5 × 7C5 + 6C6 × 7C4 = 105 + 126 + 35 = 266

## Example 3

**A boy has 3 library tickets and 8 books of his interest in the library. Of these 8, he does not want to borrow Mathematics Part II, unless Mathematics Part I is also borrowed. Then, in how many ways can he choose the three books to be borrowed?**

**Ans.** Let us make the following cases:

According to the combination formula,

Case (i)

Boy borrows Mathematics Part II, then he borrows Mathematics Part I also.

So the number of possible choices is 6C1

= 6.

Case (ii)

Boy does not borrow Mathematics Part II, then the number of possible choices is 7C3 = 35.

Hence, the total number of possible choices is 35 + 6 = 41.

## Example 4

**In how many ways can a committee consisting of 3 men and 2 women choose from 7 men and 5 women?**

**(A) 45 **

**(B) 350 **

**(C) 4200 **

**(D) 230**

**Ans.** (B) is the correct choice.

According to the combination formula, out of 7 men, 3 men can be chosen in 7C3 ways, and out of 5 women, 2 women can be chosen in 5C2 ways.

Hence, the committee can be chosen in 7C3 × 5C2 = 350 ways.

## Example 5

**In a small village, there are 87 families, of which 52 families have almost 2 children. Then, in the same way a rural development program, 20 families are to be chosen for assistance, of which at least 18 families must have at most 2 children. Now, in how many ways can the choice be made?**

**Ans.** Given that 52 of the 87 households have no more than two children, the remaining 35 families are of various types. However, according to the question, 20 households will be picked for aid under the rural development programme, with at least 18 of them having no more than two children. As a result, there are a variety of options available:

35C2 × 52C18 (18 families having at most 2 children and 2 selected from other types of families)

Currently, 52C19 × 35C1 (19 families having at most 2 children and 1 selected from other types of families)

In the same way, 52C20 (All selected 20 families having at most 2 children)

As a result, the total number of options, according to the combination formula, is

52C18 × 35C2 + 52C19 × 35C1 + 52C20

**Frequently Asked Questions about Combination Formula **

### How do you calculate combinations?

**Ans. **nCr = n! / r!×(n – r)! Is the formula for solving problems.

### What is a combination formula?

**Ans. **A combination formula is a mathematical technique for calculating the number of possible arrangements in a set of things.