Explainer: Properties of Combinations

In this explainer, we will learn how to solve problems involving combinations.

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 permutationhy 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: Charlotte, Michael, Victoria, James, and Sophia. If we choose Charlotte, then Sophia, this would not be the same as Sophia, then Charlotte 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 Charlotte, then Sophia or Sophia, then Charlotte. 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!=𝑛!𝑟!(𝑛𝑟)!.

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=𝑛!𝑟!(𝑛𝑟)!.

Substituting 𝑛=23 and 𝑟=8, we have 𝐶=23!8!(238)!=23!8!×15!.

Similarly, substituting in 𝑛=23 and 𝑟=6, we have 𝐶=23!6!×17!.

Substituting these into the given expression, we get !!×!!!×!𝐶𝐶=.

Using rules of fractions, we can rewrite this as 𝐶𝐶=23!8!×15!6!×17!23!.

Canceling the common factor of 23!, we have 𝐶𝐶=6!×17!8!×15!.

Since 𝑛!=𝑛(𝑛1)××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.

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

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

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=𝑛.

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=𝑛!𝑟!(𝑛𝑟)! that there is a symmetry about the denominator. If we substitute 𝑛𝑟 for 𝑟 in the formula, we find that we get the same expression: 𝐶=𝑛!(𝑛𝑟)!(𝑛(𝑛𝑟))!=𝑛!𝑟!(𝑛𝑟)!.

Thus, 𝐶=𝐶.

This has some interesting implications for solving equations involving 𝑃 with knowns 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=𝑛𝑏.

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 𝑛42=43 or 𝑛42=𝑛43. Since the latter of these is inconsistent, we have that the only solution is 𝑛=85.

Example 6: Solving Combinations Problems

Given that 3×𝐶,4×𝐶,6×𝐶, form an arithmetic sequence and 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×𝐶4×𝐶=4×𝐶3×𝐶.

Rearranging, we get 8×𝐶=6×𝐶+3×𝐶.

Using the definition 𝐶=𝑛!𝑟!(𝑛𝑟)!, we can rewrite this as 8𝑛!11!(𝑛11)!=6𝑛!12!(𝑛12)!+3𝑛!10!(𝑛10)!.

Dividing by the common factor of 𝑛!, we have 811!(𝑛11)!=612!(𝑛12)!+310!(𝑛10)!.

We can now multiply through by 12!(𝑛10)! to get 8×12!(𝑛10)!11!(𝑛11)!=6×12!(𝑛10)!12!(𝑛12)!+3×12!(𝑛10)!10!(𝑛10)!.

Using the property of the factorial that 𝑛!=𝑛(𝑛1)!, we can rewrite this as 8×12×11!11!×(𝑛10)(𝑛11)!(𝑛11)!=6×12!12!×(𝑛10)(𝑛11)(𝑛12)!(𝑛12)!+3×12×11×10!10!×(𝑛10)!(𝑛10)!.

Canceling the common factors in the numerators and denominators, we have 8×12(𝑛10)=6(𝑛10)(𝑛11)+3×12×11.

We can now divide through by 6 to get 8×2(𝑛10)=(𝑛10)(𝑛11)+3×2×11.

Expanding the parentheses, we get 16𝑛160=𝑛21𝑛+110+66.

By gathering like terms, we arrive at the quadratic 0=𝑛37𝑛+336𝑚.

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: 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=(𝑛1)!𝑟!(𝑛𝑟1)!+(𝑛1)!(𝑟1)!(𝑛𝑟)!.

We would like to express this as a single fraction over the common denominator of 𝑟!(𝑛𝑟)!. We can do this by multiplying the first term by 𝑛𝑟𝑛𝑟 and the second term by 𝑟𝑟 as follows: 𝐶+𝐶=(𝑛𝑟)(𝑛1)!𝑟!(𝑛𝑟)(𝑛𝑟1)!+𝑟(𝑛1)!𝑟(𝑟1)!(𝑛𝑟)!.

Using the properties of factorials that 𝑛!=𝑛(𝑛1)!, we can rewrite this as 𝐶+𝐶=(𝑛𝑟)(𝑛1)!𝑟!(𝑛𝑟)!+𝑟(𝑛1)!𝑟!(𝑛𝑟)!.

Expressing this as a single fraction and expanding the parentheses, we have 𝐶+𝐶=𝑛(𝑛1)!𝑟(𝑛1)!+𝑟(𝑛1)!𝑟!(𝑛𝑟)!.

Simplifying and using the same rule of factorials, we have 𝐶+𝐶=𝑛!𝑟!(𝑛𝑟)!=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=𝑛!𝑟!(𝑛𝑟)!, we can rewrite this expression as 𝐶+𝐶+𝐶++𝐶=5!0!×5!+5!1!×4!+5!2!×3!++5!5!×0!.

Evaluating each expression, 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 as 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)=2𝐶.

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 𝐶𝐶+𝐶𝐶+𝐶.


Recall that 𝐶=𝑛!𝑟!(𝑛𝑟)!. Using this, we see that 𝐶𝐶+𝐶𝐶+𝐶=4!0!×4!4!1!×3!+4!2!×2!4!3!×1!+4!4!×0!.

Evaluating each term gives us 𝐶𝐶+𝐶𝐶+𝐶=14+64+1=0.

Once again, this is a general rule, that the alternating sums of 𝐶 are zero: 𝐶𝐶+𝐶+(1)(𝐶)+(1)(𝐶)=0 or, more concisely, (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.

Key Points

  1. The number of combinations of size 𝑟 taken from a set of size 𝑛 is given by 𝐶=𝑃𝑟!=𝑛!𝑟!(𝑛𝑟)!.
  2. Combinations have the following key properties:
    • Reflective property: 𝐶=𝐶,
    • Recursive property: 𝐶+𝐶=𝐶,
    • Summation: 𝐶=2,
    • Alternating sum: (1)(𝐶)=0.
  3. Using the definition of 𝐶 and its properties, we can simplify many expressions and solve equations involving combinations.

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