Combinations Formula with Examples

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 the 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)!

Formula for Combinations

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!
⇒ RHS = [ n!/ r!(n-r)!]r!
⇒ RHS = n!/(n-r)! = ⁿP_r

Hence, the theorem is verified.

Read More about Difference between Permutation and Combination.

Key Points to Remember for Combination Formula

• 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_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

Read More,

• Permutation and Combination
• Aptitude Questions for Permutation and Combination
• Tricks to Solve Permutation & Combination Questions

Sample Questions

Question 1: 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

³⁰C₄ = 30! / (4!(30-4)!)
^{⇒ 30}C₄ = 30! / (4!26!)
^{⇒ 30}C₄ = 27,405 ways

Question 2: 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 3: 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
⇒ ⁸C₂ = 8!/ (2! 6!)
⇒ ⁸C₂ = 8 ×7 / (2 × 1)
⇒ ⁸C₂ = 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 4: 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

Thus, 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