# Combinations Formula with Examples

• Last Updated : 22 Sep, 2022

Combination is a way of selecting items from a collection of items in combination 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. For example- suppose that we have three numbers say, a, b, and c. Then in how many ways we can select two numbers is known as a combination.

### 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 totalof  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?

The 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.

### Combinations

The 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 denoted by nCr 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., nPr is equal to nCr × r!

Hence,                                                                     nPr = nCr × r! , 0≤ r ≤n.

This implies,                                                             nCr =   n! ⁄ r! (n-r)!

### Combination Formula

The number of combinations of n different things taken r at a time is given by

nCr = 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 Combination Formula and Permutation 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: nPr = nCr × r!

Proof:

Consider RHS, nCr × r! =[ n!/ r!(n-r)!]r!

=n!/(n-r)! = nPr

Hence, the theorem is verified.

Remarks:

1. We have nCr = n!/r!(n-r)! In particular, if r = n, then nCn = n! /n! = 1
2. nC0 = n! /0! (n-0)! = n!/0!n! = 1/0! = 1. Thus the formula nCr = n!/r!(n-r)! is applicable for r = 0 also. Hence, nCr = n!/r!(n-r)! , 0 ≤ r ≤ n
3. nCr = 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, nCr = n(n-1)(n-2)…….r factors/ r!
4. nCn-r = n!/ (n-r)![n-(n-r)]! = n!/ (n-r)! r! = nCr. Hence, nCr = nCn-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 × 29 × 28)/ (4 × 3 × 2 × 1)

= 24360/ 24

= 1015 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

and 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, 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 colour

= 6C3  × 5C3 × 7C3

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

= 20 × 10 × 35

= 7000 ways

My Personal Notes arrow_drop_up