Algebra 2 - Combinations


  • A combination is an arrangement of objects taken from n distinct objects without reference to the order. 
  • The number of combinations of n objects taken r at a time is expressed as Crn=n!r!n-r! or nr, Cn,r
  • For instance, the combination of letters m, n, x, y taken three (3) at a time are m, n, x, m, n, y, m, x, y, n, x, y.
  • Observe that the following arrangement of letters are the same: m, n, x, x, n, m, n, m, x, x, m, n, n, x, m, m, x, n. Each of these arrangements simply denotes m, n, x.
  • The notation Crn is read as "Combination n taken r".
  • In Binomial Theorem, the concept of Combinations is used. To recap, we say that the combination formula nr=nn-1n-2...n-r+11·2·3...r-1r where both r and n are positive integers and rn generates the binomial coefficients in an expansion.

Relationship Between Permutation and Combination

Suppose that we have 9 numbers in code 746395218. If we determine the number of ways to rearrange (permute) any three numbers from it, we compute it as 9·8·7=504

Assume that order is not important and we are only interested in the 3 numbers 759 being selected from the code 746395218. If an order does not matter, we count the permutations 759, 795, 957, 597, 579, 975 as one combination.

How are the combinations of 9 things taken 3 at a time related to their permutations?

For any one combination consisting of 3 numbers, say 759, there are 3! permutations. Hence, we form a relationship that Number of Combinations=Number of Permutations3!

To illustrate, we have: Number of Combinations=9·8·73!=84


Example 1. How many combinations are there in 5 distinct things taken 4 at a time?

Solution: Applying the formula gives:



Example 2. A class has 5 boys and 7 girls. In how many ways can the class elect the president, the vice president, the secretary, the treasurer, and the auditor? In how many ways can the class elect 5 members of a certain committee?

Solution 1. The first question is related to permutations since electing the said positions implies distinct filing. 

P512=12!12-5!=95,040 ways

Solution 2. The second question is related to combinations because electing 5 members of a certain committee implies one position only.

C512=12!5!12-5!=792 ways


Example 3. In how many ways can a student answer 10 out of 25 questions if he/she is required to answer 5 of the first 9 questions?


Since 5 of the 9 questions must be answered, we have C9,5 ways.

The student must now answer 5 of the remaining 16 questions, which can be done in C16,5 ways.

C9,5·C16,5=550,368 ways


Example 4. From a group of 9 men and 7 women, six persons are to be selected to form a committee so that at least four men are there on the committee. In how many ways can this be done?


We have (4 men and 2 women), (5 men and 1 woman), and (6 men and 0 women).

C9,4·C7,2+C9,5·C7,1+C9,6·C7,0=3612 ways


Cheat Sheet

  • Use the formula C=n!r!n-r! when order does not matter, and repetition is allowed.
  • Use the formula C=n+r-1!r!n-1! when order is not important, and repetition is not allowed.
  • To determine the sum of all possible combinations of n distinct things, always bear in mind the relationship Cn,0+Cn,1+Cn,2+Cn,3+...+Cn,n=2n
  • To determine the number of diagonals that can be drawn in a polygon, we use the formula Cn,2-n where n=number of sides and the constant 2 denotes two endpoints of a diagonal.
  • To find the number of combinations taking n different elements 1 at a time, 2 at a time, and so on up to n at a time, we use C=2n-1.

Blunder Areas

  • Always analyze a problem situation to make sure whether the order of arrangement matters or not. This determines whether the problem is related to combinations or permutations.