Lesson Explainer: Properties of Combinations Mathematics

In this explainer, we will learn how to use the properties of combinations to simplify expressions and solve equations.

A combination is a selection of 𝑟 items chosen without repetition from a collection of 𝑛 items in which order does not matter. The key difference between a combination and a permutation is the idea that order does not matter. For a permutation, order matters. Consider counting the number of ways we can assign the role of president and vice president to a group of 5 people: Mona, Amer, Samar, Bassem, and Dalia. If we choose Mona, then Dalia, this would not be the same as Dalia, then Mona since the first choice would be the president and the second the vice president. However, if we just wanted a committee of two people, it would not matter if we chose Mona, then Dalia or Dalia, then Mona. Hence, counting with permutations results in us overcounting the number of possible choices if order does not matter. In fact, we overcount by a factor of exactly 𝑟. Therefore, we can define the number of 𝑟 combinations from 𝑛 as the number of 𝑟 permutations of 𝑛 divided by 𝑟.

Definition: Number of Combination of a Given Size

The number of combinations of size 𝑟 taken from a collection of 𝑛 items is given by 𝐶=𝑃𝑟=𝑛𝑟𝑛−𝑟.

The notation 𝐶 can be read as 𝑛-𝐶-𝑟 or as 𝑛 choose 𝑟 and is also referred to as the binomial coefficient. Another extremely common notation for 𝐶 is 𝑛𝑟; however, there are also various other forms of notation commonly used such as 𝐶, 𝐶, 𝐶, and 𝐶(𝑛,𝑟).

This explainer will focus on the key properties of 𝐶 and how we can apply these to simplify expressions and solve equations. We begin by looking at an example where we use the formula to evaluate an expression involving combinations.

Example 1: Evaluating Combinations

Determine the value of 𝐶𝐶 without using a calculator.


Recall that 𝐶=𝑛𝑟𝑛−𝑟.

Substituting 𝑛=23 and 𝑟=8, we have 𝐶=23823−8=238×15.

Similarly, substituting in 𝑛=23 and 𝑟=6, we have 𝐶=236×17.

Substituting these into the given expression, we get ××𝐶𝐶=2381523617.

Using rules of fractions, we can rewrite this as 𝐶𝐶=238×156×1723.

Canceling the common factor of 23, we have 𝐶𝐶=6×178×15.

Since 𝑛=𝑛(𝑛−1)×⋯×2×1, we can simplify this further to get 17×168×7=347.

To solve the previous example, we could have simply used the combinations function on our calculator to evaluate the expression. However, growing in fluency in manipulating the formulae for permutations and combinations will give us the necessary skills we need to tackle more challenging problems.

Let’s consider an example where we find an unknown from an equation that involves a permutation and a combination.

Example 2: Equality of Combinations and Permutations

If 𝐶=𝑃, find the value(s) of 𝑟.


Recall, from the definition of combinations, that we have 𝐶=𝑃𝑟.

Substituting this into the given equation results in 𝑃=𝐶=𝑃𝑟.

Cross multiplying by 𝑟 and dividing by 𝑃, we can rewrite this as 𝑟=1.

We might be temped to immediately jump to the conclusion that 𝑟=1. However, this would only be a partial answer since, recalling the definition of the factorial, we also have that 0=1.

Notice that when 𝑟=0, we have 𝐶=𝑃=1 and when 𝑟=1, we have 𝐶=𝑃=𝑛.

Hence, the two possible values of 𝑟 are 1 and 0.

In the next example, we will find the expression involving permutations which is equal to a given expression involving combinations.

Example 3: Relationship between Combinations and Permutations

Which of the following is equal to 𝐶×𝐶𝐶?

  1. 𝑃𝑃
  2. 𝑃𝑃
  3. 𝑃𝑃
  4. 𝑃𝑃


We begin by noting that 𝐶=𝑛. Hence, we can rewrite out the expression: 𝐶×𝐶𝐶=𝑛𝐶𝐶.

Since all we are trying to find is an expression involving permutations, we should try to express the combinations in terms of permutations. To do this, we can use the definition that 𝐶=𝑃𝑟 to rewrite our expression as 𝐶×𝐶𝐶=𝑛67.

Canceling 6, we have 𝐶×𝐶𝐶=7𝑛𝑃𝑃, which we can also write as 𝐶×𝐶𝐶=𝑃.

Recalling the property of permutations that 𝑛𝑃=𝑃, we can we rewrite 𝑃7𝑛=𝑃.

Hence, 𝐶×𝐶𝐶=𝑃𝑃.

Therefore, the correct answer is C.

Thus far, we have simply used the definition and formula for 𝐶 to solve problems. Many problems involving combinations can be solved this way. However, oftentimes, we can solve problems in a simpler and more straightforward manor by being familiar with the properties of combinations. One such property is related to the symmetry of combinations.

Notice from the definition of 𝐶=𝑛𝑟𝑛−𝑟 that there is a symmetry about the denominator. If we substitute 𝑛−𝑟 for 𝑟 in the formula, we find that we get the same expression: 𝐶=𝑛𝑛−𝑟𝑛−(𝑛−𝑟)=𝑛𝑟𝑛−𝑟.

This leads to the general identity for combinations.

Identity: Symmetry of Combinations

Given positive integers 𝑟 and 𝑛 satisfying 𝑟<𝑛, we have 𝐶=𝐶.

This has some interesting implications for solving equations involving 𝐶 with unknowns in 𝑟. The next example will demonstrate one such implication.

Example 4: Symmetry of Combinations

Find the possible values of 𝑟 which satisfy the equation 𝐶=𝐶.


Using the rule 𝐶=𝐶, we get that 𝐶=𝐶=𝐶.

Thus, 𝑟=15 or 𝑟=6.

The last example demonstrated that if 𝐶=𝐶 then 𝑎=𝑏 or 𝑎=𝑛−𝑏.

Let’s consider another example that requires symmetry of combinations.

Example 5: Using the Symmetry of Combinations

If 𝐶+𝐶=2𝐶, find 𝑛.


Using the property that 𝐶=𝐶, we can rewrite 𝐶=𝐶. Substituting this into the given equation, we find 2𝐶=2𝐶.

This implies that 𝑛−42=43 or 𝑛−42=𝑛−43. Since the latter of these is inconsistent, we have that the only solution is 𝑛=85.

In the next example, we will determine an unknown constant in combinations when we are given that the expressions involving combinations form an arithmetic sequence.

Example 6: Solving Combinations Problems

Given that 3×𝐶,4×𝐶,6×𝐶, is an arithmetic sequence, find all possible values of 𝑛.


In an arithmetic sequence, there is a constant difference between consecutive terms. Hence, the difference between the first two and last two terms will be equal and we can write 6×𝐶−4×𝐶=4×𝐶−3×𝐶.

Rearranging, we get 8×𝐶=6×𝐶+3×𝐶.

Using the definition 𝐶=𝑛𝑟𝑛−𝑟, we can rewrite this as 8𝑛11𝑛−11=6𝑛12𝑛−12+3𝑛10𝑛−10.

Dividing by the common factor of 𝑛, we have 811𝑛−11=612𝑛−12+310𝑛−10.

We can now multiply through by 12𝑛−10 to get 8×12𝑛−1011𝑛−11=6×12𝑛−1012𝑛−12+3×12𝑛−1010𝑛−10.

Using the property of the factorial that 𝑛=𝑛𝑛−1, we can rewrite this as 8×12×1111×(𝑛−10)𝑛−11𝑛−11=6×1212×(𝑛−10)(𝑛−11)𝑛−12𝑛−12+3×12×11×1010×𝑛−10𝑛−10.

Canceling the common factors in the numerators and denominators, we have 8×12(𝑛−10)=6(𝑛−10)(𝑛−11)+3×12×11.

We can now divide through by 6 to get 8×2(𝑛−10)=(𝑛−10)(𝑛−11)+3×2×11.

Expanding the parentheses, we get 16𝑛−160=𝑛−21𝑛+110+66.

By gathering like terms, we arrive at the quadratic 0=𝑛−37𝑛+336.

Solving this by factoring or by the quadratic formula yields 𝑛=21 and 𝑛=16.

One of the other key properties of combinations is the recursive relationship:

Formula: Recursive Relationship in Combinations

𝐶+𝐶=𝐶, where 0<𝑟<𝑛.

To derive this formula, we can use the definition 𝐶 to write the left-hand side as 𝐶+𝐶=𝑛−1𝑟𝑛−𝑟−1+𝑛−1𝑟−1𝑛−𝑟.

We would like to express this as a single fraction over the common denominator of 𝑟𝑛−𝑟. We can do this by multiplying the first term by 𝑛−𝑟𝑛−𝑟 and the second term by 𝑟𝑟 as follows: 𝐶+𝐶=(𝑛−𝑟)𝑛−1𝑟(𝑛−𝑟)𝑛−𝑟−1+𝑟𝑛−1𝑟𝑟−1𝑛−𝑟.

Using the properties of factorials that 𝑛=𝑛𝑛−1, we can rewrite this as 𝐶+𝐶=(𝑛−𝑟)𝑛−1𝑟𝑛−𝑟+𝑟𝑛−1𝑟𝑛−𝑟.

Expressing this as a single fraction and expanding the parentheses, we have 𝐶+𝐶=𝑛𝑛−1−𝑟𝑛−1+𝑟𝑛−1𝑟𝑛−𝑟.

Simplifying and using the same rule of factorials, we have 𝐶+𝐶=𝑛𝑟𝑛−𝑟=𝐶 as required.

We will now turn our attention to one example where we apply this property to simplify an equation.

Example 7: Pairwise Sums of Combinations

Determine the value of 𝐶+𝐶.


This expression looks like it will be extremely laborious to evaluate or difficult to simplify. However, the first insight we gain is through noticing that when 𝑟=15 in the sum, we have the term 𝐶=𝐶. Taking this term out of the summation, we have

At this point, we can apply the recursive relationship, 𝐶+𝐶=𝐶, and simplify this to 𝐶+𝐶=𝐶+𝐶.

Now we see that if we do the same thing again and take the last term out of the summation, we have

Hence, 𝐶+𝐶=𝐶+𝐶.

Continuing the same method, we will eventually get to the last term in the sum, 𝐶, and have the expression

Therefore, the whole expression simplifies to 𝐶+𝐶=𝐶.

For the final couple of examples, we will consider the sums of all combinations 𝐶 for a given 𝑛.

Example 8: Sums of Combinations

Find the value of 𝐶+𝐶+𝐶+⋯+𝐶.


Using the definition 𝐶=𝑛𝑟𝑛−𝑟, we can rewrite this expression as 𝐶+𝐶+𝐶+⋯+𝐶=50×5+51×4+52×3+⋯+55×0.

Evaluating each term, we have 𝐶+𝐶+𝐶+⋯+𝐶=1+5+10+10+5+1=32.

In the last example, we found that the sum of all combinations 𝐶 for 𝑛=5 is 32; it is no coincidence that this is equal to 2. In fact, the general rule is that the sum of all 𝐶 for any given 𝑛 is equal to 2. We can write this as 𝐶+𝐶+𝐶+⋯+𝐶+𝐶=2, or more concisely, we have the following identity.

Identity: Summation of Combinations

For any positive integer 𝑛, we have 𝐶=2.

This rule is maybe not so surprising when we consider the recursive relationship for each term: 𝐶+𝐶=𝐶.

Since this does not apply for 𝑟=0 or 𝑟=𝑛, we can rewrite the sum as 𝐶=𝐶+𝐶+𝐶+𝐶+𝐶+𝐶+𝐶+⋯+𝐶+𝐶+𝐶.

Since 𝐶=1=𝐶 and 𝐶=1=𝐶, we can rewrite this expression as 𝐶=𝐶+𝐶+𝐶+𝐶+𝐶+𝐶+𝐶+⋯+𝐶+𝐶+𝐶.

Regrouping the terms, we have 𝐶=𝐶+𝐶+𝐶+𝐶+𝐶+𝐶+⋯+𝐶+𝐶+𝐶+𝐶=2𝐶.

Therefore, the sum of the 𝐶 is twice the sum of 𝐶. Moreover, since 𝐶=1, we can see that the sum of 𝐶 for a given 𝑛 will be a power of two; in particular, it will be 2.

Finally, we will consider the alternating sum of combinations.

Example 9: Alternating Sums of Combinations

Find the value of 𝐶−𝐶+𝐶−𝐶+𝐶.


Recall that 𝐶=𝑛𝑟𝑛−𝑟. Using this, we see that 𝐶−𝐶+𝐶−𝐶+𝐶=40×4−41×3+42×2−43×1+44×0.

Evaluating each term gives us 𝐶−𝐶+𝐶−𝐶+𝐶=1−4+6−4+1=0.

Once again, this is a general rule, that the alternating sums of 𝐶 are zero: 𝐶−𝐶+𝐶−⋯+(−1)𝐶+(−1)𝐶=0 or, more concisely,

Identity: Alternating Sum of Combinations

For any positive integer 𝑛, we have (−1)𝐶=0.

An alternative way to represent this is that the sums of odd and even terms are equal. This is not surprising when 𝑛 is odd due to the reflective symmetry: 𝐶=𝐶. However, as the previous example showed, this is also true for even 𝑛.

Let’s recap a few important concepts from the explainer.

Key Points

  • The number of combinations of size 𝑟 taken from a set of size 𝑛 is given by 𝐶=𝑃𝑟=𝑛𝑟𝑛−𝑟.
  • Combinations have the following key properties: given positive integers 𝑟 and 𝑛 satisfying 𝑟<𝑛,
    • Symmetry property: 𝐶=𝐶,
    • Recursive property: 𝐶+𝐶=𝐶,
    • Summation: 𝐶=2,
    • Alternating sum: (−1)𝐶=0.
  • Using the definition of 𝐶 and its properties, we can simplify many expressions and solve equations involving combinations.

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