Combinations are way of selecting items from a collection of items in combinations we do not look at the order of selecting items, but our main attention is on the total number of selected items from a given set of items.

Combinations formula is main concept for figuring out how many ways you can choose items from a larger group, where the order doesn’t matter. This is important in many areas like math, statistics, computer science, and finance. The combinations formula, written as C(n,k)C(n, k)C(n,k) or (nk)\binom{n}{k}(kn​), helps you calculate the number of possible combinations. This Combinations formula is useful for solving problems related to probability, planning, and making decisions. In this article, we will explain the combination formula in simple terms, show how it works, and give examples of how to use it in real-life situations

What is Combinations?

Combinations are way of selecting items from a collection of items. Different groups that can be formed by choosing r things from a given set of n different things, ignoring their order of arrangement, are called combinations of n things taken r at a time.

The number of all such combinations is calculated by combinations formula : ⁿC_r or C(n, r).

Example: All the combinations of four different objects a, b, c, d taken two at a time are ab, ac, ad, bc, bd, cd. Here we have not included ba, ca, da, cb, db, and dc as the order does not alter the combination. Thus there are 6 combinations of 4 different objects taken 2 at a time i.e., ⁴C₂ = 6

Similarly, all the combinations of four different objects a, b, c and d taken three at a time are abc, bcd, cda, dab.

Thus there are four combinations of 4 different objects taken 3 at a time i.e., ⁴C₃ = 4

Corresponding to each of these combinations, we have 3! permutations, as three objects in each combination can be arranged among themselves in 3! ways. Hence, the number of permutations 

= ⁴C₃ × 3!

⁴P₃ = ⁴C₃ × 3!

4!/(4-3)! 3! = ⁴C₃

Thus we can conclude that the total number of permutations of n different things taken r at a time i.e., ⁿP_r is equal to ⁿC_r × r!

Hence,

ⁿP_r = ⁿC_r × r! , 0 ≤ r ≤ n

This implies,

ⁿC_r =   n! ⁄ r! (n-r)!

Combinations Formula

The combinations formula provides a way to calculate the number of combinations of n different things taken r at a time is given by

ⁿC_r = n! ⁄  r! (n-r)! ,0 < r ≤n

where,

• n is the size of the set from which elements are permuted
• r is the size of each permutation
• ! is factorial operator

Relation between Combinations Formula and Permutations Formula

The main difference between combination and permutation is only that in permutation we also consider the order of selecting the things but in combination order of selection does not matter. And therefore, permutations are always greater than the combination.

Theorem: ⁿP_r = ⁿC_r × r! ( Permutations formula = Combinations formula × r! )

Proof:

Consider,

RHS, ⁿC_r × r!

= [ n!/ r!(n-r)!]r!

= n!/(n-r)! = ⁿP_r

Hence, the theorem is verified.

Difference between Permutations and Combinations

Each of the arrangements that can be made out of a given set of things, by taking some or all of them at a time, are called Permutations. The order in which arrangements are taken is important in a Permutation.

Each of the groups or selections (in any order) that can be made out of a given set of things by taking some or all of them at a time are called combinations. The order in which selections are made is not important in a Combination.

Example: Two letters a and b together form one group (combination), but they can be arranged in two different ways as ab and ba and thus there are total of  two arrangements (permutations).

Again, if we take three letters a, b, and c, then the number of groups taking two letters at a time is three i.e., ab, bc, and ca.

But each group gives rise to two different arrangements, hence the total number of arrangements = 6 i.e., ab, ba, bc, cb, ca, and ac.

Further, if we take four letters a, b, c, and d, then the combinations which can be made by taking two letters at a time are six in numbers.

ab, ac, ad, bc, bd, cd

And the permutations which can be made by taking two letters at a time are twelve in number

ab, ba, ac, ca, ad, da, bc, cb, bd, db, cd, dc

Before we proceed on to study permutations and combinations in detail, we shall introduce a notation n! read as n factorial, which is very helpful in the study and calculation of permutations and combinations.

What is Factorial?

Continued product of first n natural numbers (i.e., the product of 1, 2, 3, …, n) is denoted by symbol n! and read as factorial n.

For example, 5! = 1.2.3.4.5 = 120

In general,

n! = 1.2.3.4…..(n – 1).n

Note:

1. We define 1! = 1 and 0! = 1
2. n! is not defined when n is a negative integer or a fraction.

Remarks:

• We have ⁿC_r = n!/r!(n-r)! In particular, if r = n, then ⁿC_n = n! /n! = 1
• ⁿC₀ = n! /0! (n-0)! = n!/0!n! = 1/0! = 1. Thus the formula ⁿC_r = n!/r!(n-r)! is applicable for r = 0 also. Hence, ⁿC_r = n!/r!(n-r)! , 0 ≤ r ≤ n
• ⁿC_r = n! /r!(n-r)! = n(n-1)(n-2)……..(n-r+-1)(n-r)(n-r-1)…….3.2.1 / r! [(n-r)(n-r-1)…..3.2.1]. Therefore, ⁿC_r = n(n-1)(n-2)…….r factors/ r!
• ⁿC_n-r = n!/ (n-r)![n-(n-r)]! = n!/ (n-r)! r! = ⁿC_r. Hence, ⁿC_r = ⁿC_n-r i.e., selecting r objects is same as rejecting (n-r) objects

Sample Questions

Question 1: Evaluate 4! – 3!

Solution:

4! – 3!

= (4 × 3 × 2 × 1) – (3 × 2 × 1)

= 24 – 6

=18

Question 2: From a class of 30 students, 4 are to be chosen for the competition. In how many ways can they be chosen?

Solution:

Total students = n = 30

Number of students to be chosen = r = 4

Hence, Total number of ways 4 students out of 30 can be chosen is,

³⁰C₄

= 30! / (4!(30-4)!)

= 30! / (4!26!)

= 27,405 ways

Question 3: Nitin has 5 friends. In how many ways can he invite one or more of them to his party.

Solution:

Nitin may invite (i) one of them (ii) two of them (iii) three of them (iv) four of them (v) all of them 

with combinations formula we can calculate that this can be done in ⁵C₁, ⁵C₂, ⁵C₃, ⁵C₄, ⁵C₅ ways

Therefore, The total number of ways

= ⁵C₁ +  ⁵C₂ + ⁵C₃ + ⁵C₄ +  ⁵C₅

= 5!/ (1! 4!) + 5!/ (2! 3!) + 5!/ (4! 1!) + 5!/ (5! 0!)

=  5 + 10 + 10 + 5 +1

= 31 ways

Question 4: Find the number of diagonals that can be drawn by joining the angular points of an octagon.

Solution:

A diagonal is made by joining any two angular points. 

There are 8 vertices or angular points in an octagon

Therefore, by using combinations formula the Number of straight lines formed

= ⁸C₂ = 8!/ (2! 6!)

= 8 ×7 / (2 × 1)

= 56/ 2

= 28

Which also includes the 8 sides of the octagon

Therefore, Number of diagonal

= 28 – sides of octagon

= 28 – 8

= 20 diagonals

Question 5: Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls, and 7 blue balls, if each selection consists of 3 balls of each color.

Solution:

Number of Red balls = 6

Number of white balls = 5

Number of Blue balls = 7

Total number of balls to be selected = 9

Hence, the required number of ways of selecting 9 balls from 6 red, 5 white, 7 blue balls consisting of 3 balls of each color can be calculated by using combinations formula :

= ⁶C₃  × ⁵C₃ × ⁷C₃

= 6!/ (3! 3!) × 5!/ (3! 2!) × 7!/ (3! 4!)

= 20 × 10 × 35 

= 7000 ways

Combinations Formula- FAQs

When should I use the combinations formula?

You can use the combinations formula when you need to determine the number of ways to select a group of items from a larger set where the order of selection does not matter. Examples include selecting committee members, forming teams, or picking lottery numbers.

What is the value of (ⁿC₀) and (ⁿC_n)?

The value of (ⁿC₀) is 1 because there is exactly one way to choose zero items from a set. Similarly, (ⁿC_n) is also 1 because there is exactly one way to choose all n items from the set.

What is a factorial?

A factorial, denoted by !, is the product of all positive integers up to a given number. For example, 5!=5×4×3×2×1=120.

How do you calculate combinations using the formula?

To calculate combinations:

1. Identify n (the total number of items) and k (the number of items to choose).
2. Substitute these values into the combinations formula: C(n, k) = {n!}/{k!(n – k)!}
3. Calculate the factorials and simplify the expression to find the number of combinations.