Explainer: Properties of Combinations

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

A combination is a selection of ๐‘Ÿ items chosen without repetition from a collection of ๐‘› 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 ๐‘Ÿ!. Therefore, we can define the number of ๐‘Ÿ combinations from ๐‘› as the number of ๐‘Ÿ permutations of ๐‘› divided by ๐‘Ÿ!.

Definition: Number of Combination of a Given Size

The number of combinations of size ๐‘Ÿ taken from a collection of ๐‘› items is given by ๏Š๏Ž๏Š๏Ž๐ถ=๐‘ƒ๐‘Ÿ!=๐‘›!๐‘Ÿ!(๐‘›โˆ’๐‘Ÿ)!.

The notation ๏Š๏Ž๐ถ can be read as ๐‘›-๐ถ-๐‘Ÿ or as ๐‘› choose ๐‘Ÿ and is also referred to as the binomial coefficient. Another extremely common notation for ๏Š๏Ž๐ถ is ๏€ฟ๐‘›๐‘Ÿ๏‹; however, there are also various other forms of notation commonly used such as ๏Š๏Ž๐ถ, ๐ถ๏Š๏Ž, ๐ถ๏Š๏Ž•๏Ž, and ๐ถ(๐‘›,๐‘Ÿ).

This explainer will focus on the key properties of ๏Š๏Ž๐ถ 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 ๏Šจ๏Šฉ๏Šฎ๏Šจ๏Šฉ๏Šฌ๐ถ๐ถ without using a calculator.

Answer

Recall that ๏Š๏Ž๐ถ=๐‘›!๐‘Ÿ!(๐‘›โˆ’๐‘Ÿ)!.

Substituting ๐‘›=23 and ๐‘Ÿ=8, we have ๏Šจ๏Šฉ๏Šฎ๐ถ=23!8!(23โˆ’8)!=23!8!ร—15!.

Similarly, substituting in ๐‘›=23 and ๐‘Ÿ=6, we have ๏Šจ๏Šฉ๏Šฌ๐ถ=23!6!ร—17!.

Substituting these into the given expression, we get ๏Šจ๏Šฉ๏Šฎ๏Šจ๏Šฉ๏Šฌ๏Šจ๏Šฉ!๏Šฎ!ร—๏Šง๏Šซ!๏Šจ๏Šฉ!๏Šฌ!ร—๏Šง๏Šญ!๐ถ๐ถ=๏€ป๏‡๏€ป๏‡.

Using rules of fractions, we can rewrite this as ๏Šจ๏Šฉ๏Šฎ๏Šจ๏Šฉ๏Šฌ๐ถ๐ถ=23!8!ร—15!6!ร—17!23!.

Canceling the common factor of 23!, we have ๏Šจ๏Šฉ๏Šฎ๏Šจ๏Šฉ๏Šฌ๐ถ๐ถ=6!ร—17!8!ร—15!.

Since ๐‘›!=๐‘›(๐‘›โˆ’1)ร—โ‹ฏร—2ร—1, we can simplify this further to get 17ร—168ร—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 ๏Š๏Ž๏Š๏Ž๐ถ=๐‘ƒ, find the value(s) of ๐‘Ÿ.

Answer

Recall, from the definition of combinations, that we have ๏Š๏Ž๏Š๏Ž๐ถ=๐‘ƒ๐‘Ÿ!.

Substituting this into the given equation results in ๏Š๏Ž๏Š๏Ž๏Š๏Ž๐‘ƒ=๐ถ=๐‘ƒ๐‘Ÿ!.

Cross multiplying by ๐‘Ÿ! and dividing by ๏Š๏Ž๐‘ƒ, we can rewrite this as ๐‘Ÿ!=1.

We might be temped to immediately jump to the conclusion that ๐‘Ÿ=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 ๐‘Ÿ are 1 and 0.

Notice that when ๐‘Ÿ=0, we have ๏Š๏Šฆ๏Š๏Šฆ๐ถ=๐‘ƒ=1 and when ๐‘Ÿ=1, we have ๏Š๏Šง๏Š๏Šง๐ถ=๐‘ƒ=๐‘›.

Example 3: Relationship between Combinations and Permutations

Which of the following is equal to ๏Š๏Šง๏Šฌ๏Š๏Šฌ๏Šญ๏Š๏Šญ๐ถร—๐ถ๐ถ?

  1. ๏Šฌ๏Š๏Šฌ๏Šญ๏Š๏Šฑ๏Šง๏Šญ๐‘ƒ๐‘ƒ
  2. ๏Šฌ๏Š๏Šฌ๏Šญ๏Š๏Šฑ๏Šญ๏Šฌ๐‘ƒ๐‘ƒ
  3. ๏Šฌ๏Š๏Šฌ๏Šญ๏Š๏Šฑ๏Šง๏Šฌ๐‘ƒ๐‘ƒ
  4. ๏Šฌ๏Š๏Šฌ๏Šญ๏Š๏Šฑ๏Šญ๏Šญ๐‘ƒ๐‘ƒ

Answer

We begin by noting that ๏Š๏Šง๐ถ=๐‘›. Hence, we can rewrite out the expression: ๏Š๏Šง๏Šฌ๏Š๏Šฌ๏Šญ๏Š๏Šญ๏Šฌ๏Š๏Šฌ๏Šญ๏Š๏Šญ๐ถร—๐ถ๐ถ=๐‘›๐ถ๐ถ.

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 ๏Š๏Ž๏Š๏Ž๐ถ=๐‘ƒ๐‘Ÿ! to rewrite our expression as ๏Š๏Šง๏Šฌ๏Š๏Šฌ๏Šญ๏Š๏Šญ๏Œฏ๏Šฌ!๏Œฏ๏Šญ!๐ถร—๐ถ๐ถ=๐‘›.๏Žฅ๏‘ƒ๏Žฅ๏Žฆ๏‘ƒ๏Žฆ

Canceling 6!, we have ๏Š๏Šง๏Šฌ๏Š๏Šฌ๏Šญ๏Š๏Šญ๏Šฌ๏Š๏Šฌ๏Šญ๏Š๏Šญ๐ถร—๐ถ๐ถ=7๐‘›(๐‘ƒ)๐‘ƒ, which we can also write as ๏Š๏Šง๏Šฌ๏Š๏Šฌ๏Šญ๏Š๏Šญ๏Šฌ๏Š๏Šฌ๏Œฏ๏Šญ๏Š๐ถร—๐ถ๐ถ=๐‘ƒ.๏Žฆ๏‘ƒ๏Žฆ

Recalling the property of permutations that ๐‘›(๐‘ƒ)=๐‘ƒ๏Š๏Šฑ๏Šง๏Ž๏Šฑ๏Šง๏Š๏Ž, we can we rewrite ๏Šญ๏Š๏Šญ๏Šญ๏Š๏Šฑ๏Šง๏Šฌ๐‘ƒ7๐‘›=๐‘ƒ.

Hence, ๏Š๏Šง๏Šฌ๏Š๏Šฌ๏Šญ๏Š๏Šญ๏Šฌ๏Š๏Šฌ๏Šญ๏Š๏Šฑ๏Šง๏Šฌ๐ถร—๐ถ๐ถ=๐‘ƒ๐‘ƒ.

Therefore, the correct answer is C.

Thus far, we have simply used the definition and formula for ๏Š๏Ž๐ถ 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 ๏Š๏Ž๐ถ=๐‘›!๐‘Ÿ!(๐‘›โˆ’๐‘Ÿ)! that there is a symmetry about the denominator. If we substitute ๐‘›โˆ’๐‘Ÿ for ๐‘Ÿ in the formula, we find that we get the same expression: ๏Š๏Š๏Šฑ๏Ž๐ถ=๐‘›!(๐‘›โˆ’๐‘Ÿ)!(๐‘›โˆ’(๐‘›โˆ’๐‘Ÿ))!=๐‘›!๐‘Ÿ!(๐‘›โˆ’๐‘Ÿ)!.

Thus, ๏Š๏Ž๏Š๏Š๏Šฑ๏Ž๐ถ=๐ถ.

This has some interesting implications for solving equations involving ๏Š๏Ž๐‘ƒ with knowns in ๐‘Ÿ. The next example will demonstrate one such implication.

Example 4: Symmetry of Combinations

Find the possible values of ๐‘Ÿ which satisfy the equation ๏Šจ๏Šง๏Ž๏Šจ๏Šง๏Šง๏Šซ๐ถ=๐ถ.

Answer

Using the rule ๏Š๏Ž๏Š๏Š๏Šฑ๏Ž๐ถ=๐ถ, we get that ๏Šจ๏Šง๏Ž๏Šจ๏Šง๏Šง๏Šซ๏Šจ๏Šง๏Šฌ๐ถ=๐ถ=๐ถ.

Thus, ๐‘Ÿ=15 or ๐‘Ÿ=6.

The last example demonstrated that if ๏Š๏Œบ๏Š๏Œป๐ถ=๐ถ then ๐‘Ž=๐‘ or ๐‘Ž=๐‘›โˆ’๐‘.

Example 5: Using the Symmetry of Combinations

If ๏Š๏Šช๏Šจ๏Š๏Š๏Šฑ๏Šช๏Šจ๏Š๏Šช๏Šฉ๐ถ+๐ถ=2(๐ถ), find ๐‘›.

Answer

Using the property that ๏Š๏Ž๏Š๏Š๏Šฑ๏Ž๐ถ=๐ถ, we can rewrite ๏Š๏Šช๏Šจ๏Š๏Š๏Šฑ๏Šช๏Šจ๐ถ=๐ถ. Substituting this into the given equation, we find 2(๐ถ)=2(๐ถ).๏Š๏Š๏Šฑ๏Šช๏Šจ๏Š๏Šช๏Šฉ

This implies that ๐‘›โˆ’42=43 or ๐‘›โˆ’42=๐‘›โˆ’43. Since the latter of these is inconsistent, we have that the only solution is ๐‘›=85.

Example 6: Solving Combinations Problems

Given that 3ร—๐ถ,4ร—๐ถ,6ร—๐ถ,๏Š๏Šง๏Šฆ๏Š๏Šง๏Šง๏Š๏Šง๏Šจ form an arithmetic sequence and find all possible values of ๐‘›.

Answer

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 6ร—๐ถโˆ’4ร—๐ถ=4ร—๐ถโˆ’3ร—๐ถ.๏Š๏Šง๏Šจ๏Š๏Šง๏Šง๏Š๏Šง๏Šง๏Š๏Šง๏Šฆ

Rearranging, we get 8ร—๐ถ=6ร—๐ถ+3ร—๐ถ.๏Š๏Šง๏Šง๏Š๏Šง๏Šจ๏Š๏Šง๏Šฆ

Using the definition ๏Š๏Ž๐ถ=๐‘›!๐‘Ÿ!(๐‘›โˆ’๐‘Ÿ)!, we can rewrite this as 8๐‘›!11!(๐‘›โˆ’11)!=6๐‘›!12!(๐‘›โˆ’12)!+3๐‘›!10!(๐‘›โˆ’10)!.

Dividing by the common factor of ๐‘›!, we have 811!(๐‘›โˆ’11)!=612!(๐‘›โˆ’12)!+310!(๐‘›โˆ’10)!.

We can now multiply through by 12!(๐‘›โˆ’10)! to get 8ร—12!(๐‘›โˆ’10)!11!(๐‘›โˆ’11)!=6ร—12!(๐‘›โˆ’10)!12!(๐‘›โˆ’12)!+3ร—12!(๐‘›โˆ’10)!10!(๐‘›โˆ’10)!.

Using the property of the factorial that ๐‘›!=๐‘›(๐‘›โˆ’1)!, we can rewrite this as 8ร—12ร—11!11!ร—(๐‘›โˆ’10)(๐‘›โˆ’11)!(๐‘›โˆ’11)!=6ร—12!12!ร—(๐‘›โˆ’10)(๐‘›โˆ’11)(๐‘›โˆ’12)!(๐‘›โˆ’12)!+3ร—12ร—11ร—10!10!ร—(๐‘›โˆ’10)!(๐‘›โˆ’10)!.

Canceling the common factors in the numerators and denominators, we have 8ร—12(๐‘›โˆ’10)=6(๐‘›โˆ’10)(๐‘›โˆ’11)+3ร—12ร—11.

We can now divide through by 6 to get 8ร—2(๐‘›โˆ’10)=(๐‘›โˆ’10)(๐‘›โˆ’11)+3ร—2ร—11.

Expanding the parentheses, we get 16๐‘›โˆ’160=๐‘›โˆ’21๐‘›+110+66.๏Šจ

By gathering like terms, we arrive at the quadratic 0=๐‘›โˆ’37๐‘›+336๐‘š.๏Šจ

Solving this by factoring or by the quadratic formula yields ๐‘›=21 and ๐‘›=16.

One of the other key properties of combinations is the recursive relationship: ๏Š๏Šฑ๏Šง๏Ž๏Š๏Šฑ๏Šง๏Ž๏Šฑ๏Šง๏Š๏Ž๐ถ+๐ถ=๐ถ, where 0<๐‘Ÿ<๐‘›. To derive this formula, we can use the definition ๏Š๏Ž๐ถ to write the left-hand side as ๏Š๏Šฑ๏Šง๏Ž๏Š๏Šฑ๏Šง๏Ž๏Šฑ๏Šง๐ถ+๐ถ=(๐‘›โˆ’1)!๐‘Ÿ!(๐‘›โˆ’๐‘Ÿโˆ’1)!+(๐‘›โˆ’1)!(๐‘Ÿโˆ’1)!(๐‘›โˆ’๐‘Ÿ)!.

We would like to express this as a single fraction over the common denominator of ๐‘Ÿ!(๐‘›โˆ’๐‘Ÿ)!. We can do this by multiplying the first term by ๐‘›โˆ’๐‘Ÿ๐‘›โˆ’๐‘Ÿ and the second term by ๐‘Ÿ๐‘Ÿ as follows: ๏Š๏Šฑ๏Šง๏Ž๏Š๏Šฑ๏Šง๏Ž๏Šฑ๏Šง๐ถ+๐ถ=(๐‘›โˆ’๐‘Ÿ)(๐‘›โˆ’1)!๐‘Ÿ!(๐‘›โˆ’๐‘Ÿ)(๐‘›โˆ’๐‘Ÿโˆ’1)!+๐‘Ÿ(๐‘›โˆ’1)!๐‘Ÿ(๐‘Ÿโˆ’1)!(๐‘›โˆ’๐‘Ÿ)!.

Using the properties of factorials that ๐‘›!=๐‘›(๐‘›โˆ’1)!, we can rewrite this as ๏Š๏Šฑ๏Šง๏Ž๏Š๏Šฑ๏Šง๏Ž๏Šฑ๏Šง๐ถ+๐ถ=(๐‘›โˆ’๐‘Ÿ)(๐‘›โˆ’1)!๐‘Ÿ!(๐‘›โˆ’๐‘Ÿ)!+๐‘Ÿ(๐‘›โˆ’1)!๐‘Ÿ!(๐‘›โˆ’๐‘Ÿ)!.

Expressing this as a single fraction and expanding the parentheses, we have ๏Š๏Šฑ๏Šง๏Ž๏Š๏Šฑ๏Šง๏Ž๏Šฑ๏Šง๐ถ+๐ถ=๐‘›(๐‘›โˆ’1)!โˆ’๐‘Ÿ(๐‘›โˆ’1)!+๐‘Ÿ(๐‘›โˆ’1)!๐‘Ÿ!(๐‘›โˆ’๐‘Ÿ)!.

Simplifying and using the same rule of factorials, we have ๏Š๏Šฑ๏Šง๏Ž๏Š๏Šฑ๏Šง๏Ž๏Šฑ๏Šง๏Š๏Ž๐ถ+๐ถ=๐‘›!๐‘Ÿ!(๐‘›โˆ’๐‘Ÿ)!=๐ถ 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 ๏Šฌ๏Šฌ๏Šจ๏Šง๏Šง๏Šซ๏Ž๏Šฒ๏Šง๏Šฎ๏Šง๏Šฑ๏Ž๏Šจ๏Šฆ๐ถ+๏„š๐ถ.

Answer

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 ๐‘Ÿ=15 in the sum, we have the term ๏Šฎ๏Šง๏Šฑ๏Šง๏Šซ๏Šจ๏Šฆ๏Šฌ๏Šฌ๏Šจ๏Šฆ๐ถ=๐ถ. Taking this term out of the summation, we have

At this point, we can apply the recursive relationship, ๏Š๏Šฑ๏Šง๏Ž๏Š๏Šฑ๏Šง๏Ž๏Šฑ๏Šง๏Š๏Ž๐ถ+๐ถ=๐ถ, and simplify this to ๏Šฌ๏Šฌ๏Šจ๏Šง๏Šง๏Šซ๏Ž๏Šฒ๏Šง๏Šฎ๏Šง๏Šฑ๏Ž๏Šจ๏Šฆ๏Šฌ๏Šญ๏Šจ๏Šง๏Šง๏Šช๏Ž๏Šฒ๏Šง๏Šฎ๏Šง๏Šฑ๏Ž๏Šจ๏Šฆ๐ถ+๏„š๐ถ=๐ถ+๏„š๐ถ.

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

Hence, ๏Šฌ๏Šฌ๏Šจ๏Šง๏Šง๏Šซ๏Ž๏Šฒ๏Šง๏Šฎ๏Šง๏Šฑ๏Ž๏Šจ๏Šฆ๏Šฌ๏Šฎ๏Šจ๏Šง๏Šง๏Šฉ๏Ž๏Šฒ๏Šง๏Šฎ๏Šง๏Šฑ๏Ž๏Šจ๏Šฆ๐ถ+๏„š๐ถ=๐ถ+๏„š๐ถ.

Continuing the same method, we will eventually get to the last term in the sum, ๏Šฎ๏Šฆ๏Šจ๏Šฆ๐ถ, and have the expression

Therefore, the whole expression simplifies to ๏Šฌ๏Šฌ๏Šจ๏Šง๏Šง๏Šซ๏Ž๏Šฒ๏Šง๏Šฎ๏Šง๏Šฑ๏Ž๏Šจ๏Šฆ๏Šฎ๏Šง๏Šจ๏Šง๐ถ+๏„š๐ถ=๐ถ.

For the final couple of examples, we will consider the sums of all combinations ๏Š๏Ž๐ถ for a given ๐‘›.

Example 8: Sums of Combinations

Find the value of ๏Šซ๏Šฆ๏Šซ๏Šง๏Šซ๏Šจ๏Šซ๏Šซ๐ถ+๐ถ+๐ถ+โ‹ฏ+๐ถ.

Answer

Using the definition ๏Š๏Ž๐ถ=๐‘›!๐‘Ÿ!(๐‘›โˆ’๐‘Ÿ)!, we can rewrite this expression as ๏Šซ๏Šฆ๏Šซ๏Šง๏Šซ๏Šจ๏Šซ๏Šซ๐ถ+๐ถ+๐ถ+โ‹ฏ+๐ถ=5!0!ร—5!+5!1!ร—4!+5!2!ร—3!+โ‹ฏ+5!5!ร—0!.

Evaluating each expression, we have ๏Šซ๏Šฆ๏Šซ๏Šง๏Šซ๏Šจ๏Šซ๏Šซ๐ถ+๐ถ+๐ถ+โ‹ฏ+๐ถ=1+5+10+10+5+1=32.

In the last example, we found that the sum of all combinations ๏Š๏Ž๐ถ for ๐‘›=5 is 32; it is no coincidence that this is equal to 2๏Šซ. In fact, the general rule is that the sum of all ๏Š๏Ž๐ถ for any given ๐‘› is equal to 2๏Š. We can write this as ๏Š๏Šฆ๏Š๏Šง๏Š๏Šจ๏Š๏Š๏Šฑ๏Šง๏Š๏Š๏Š๐ถ+๐ถ+๐ถ+โ‹ฏ+๐ถ+๐ถ=2, or more concisely as ๏Š๏Ž๏Šฒ๏Šฆ๏Š๏Ž๏Š๏„š๐ถ=2.

This rule is maybe not so surprising when we consider the recursive relationship for each term: ๏Š๏Šฑ๏Šง๏Ž๏Šฑ๏Šง๏Š๏Šฑ๏Šง๏Ž๏Š๏Ž๐ถ+๐ถ=๐ถ.

Since this does not apply for ๐‘Ÿ=0 or ๐‘Ÿ=๐‘›, we can rewrite the sum as ๏Š๏Ž๏Šฒ๏Šฆ๏Š๏Ž๏Š๏Šฆ๏Š๏Šฑ๏Šง๏Šฆ๏Š๏Šฑ๏Šง๏Šง๏Š๏Šฑ๏Šง๏Šง๏Š๏Šฑ๏Šง๏Šจ๏Š๏Šฑ๏Šง๏Šจ๏Š๏Šฑ๏Šง๏Šฉ๏Š๏Šฑ๏Šง๏Š๏Šฑ๏Šง๏Š๏Šฑ๏Šง๏Š๏Š๏Š๏„š๐ถ=๐ถ+(๐ถ+๐ถ)+(๐ถ+๐ถ)+(๐ถ+๐ถ)+โ‹ฏ+(๐ถ+๐ถ)+๐ถ.

Since ๏Š๏Šฆ๏Š๏Šฑ๏Šง๏Šฆ๐ถ=1=๐ถ and ๏Š๏Š๏Š๏Šฑ๏Šง๏Š๏Šฑ๏Šง๐ถ=1=๐ถ, we can rewrite this expression as ๏Š๏Ž๏Šฒ๏Šฆ๏Š๏Ž๏Š๏Šฑ๏Šง๏Šฆ๏Š๏Šฑ๏Šง๏Šฆ๏Š๏Šฑ๏Šง๏Šง๏Š๏Šฑ๏Šง๏Šง๏Š๏Šฑ๏Šง๏Šจ๏Š๏Šฑ๏Šง๏Šจ๏Š๏Šฑ๏Šง๏Šฉ๏Š๏Šฑ๏Šง๏Š๏Šฑ๏Šง๏Š๏Šฑ๏Šง๏Š๏Š๏Šฑ๏Šง๏Š๏Šฑ๏Šง๏„š๐ถ=๐ถ+(๐ถ+๐ถ)+(๐ถ+๐ถ)+(๐ถ+๐ถ)+โ‹ฏ+(๐ถ+๐ถ)+๐ถ.

Regrouping the terms, we have ๏Š๏Ž๏Šฒ๏Šฆ๏Š๏Ž๏Š๏Šฑ๏Šง๏Šฆ๏Š๏Šฑ๏Šง๏Šฆ๏Š๏Šฑ๏Šง๏Šง๏Š๏Šฑ๏Šง๏Šง๏Š๏Šฑ๏Šง๏Šจ๏Š๏Šฑ๏Šง๏Šจ๏Š๏Šฑ๏Šง๏Š๏Šฑ๏Šจ๏Š๏Šฑ๏Šง๏Š๏Šฑ๏Šจ๏Š๏Šฑ๏Šง๏Š๏Šฑ๏Šง๏Š๏Šฑ๏Šง๏Š๏Šฑ๏Šง๏Š๏Šฑ๏Šง๏Ž๏Šฒ๏Šฆ๏Š๏Šฑ๏Šง๏Ž๏„š๐ถ=(๐ถ+๐ถ)+(๐ถ+๐ถ)+(๐ถ+๐ถ)+โ‹ฏ+(๐ถ+๐ถ)+(๐ถ+๐ถ)=2๏„š๐ถ.

Therefore, the sum of the ๏Š๏Ž๐ถ is twice the sum of ๏Š๏Šฑ๏Šง๏Ž๐ถ. Moreover, since ๏Šฆ๏Šฆ๐ถ=1, we can see that the sum of ๏Š๏Ž๐ถ for a given ๐‘› will be a power of two; in particular, it will be 2๏Š.

Finally, we will consider the alternating sum of combinations.

Example 9: Alternating Sums of Combinations

Find the value of ๏Šช๏Šฆ๏Šช๏Šง๏Šช๏Šจ๏Šช๏Šฉ๏Šช๏Šช๐ถโˆ’๐ถ+๐ถโˆ’๐ถ+๐ถ.

Answer

Recall that ๏Š๏Ž๐ถ=๐‘›!๐‘Ÿ!(๐‘›โˆ’๐‘Ÿ)!. Using this, we see that ๏Šช๏Šฆ๏Šช๏Šง๏Šช๏Šจ๏Šช๏Šฉ๏Šช๏Šช๐ถโˆ’๐ถ+๐ถโˆ’๐ถ+๐ถ=4!0!ร—4!โˆ’4!1!ร—3!+4!2!ร—2!โˆ’4!3!ร—1!+4!4!ร—0!.

Evaluating each term gives us ๏Šช๏Šฆ๏Šช๏Šง๏Šช๏Šจ๏Šช๏Šฉ๏Šช๏Šช๐ถโˆ’๐ถ+๐ถโˆ’๐ถ+๐ถ=1โˆ’4+6โˆ’4+1=0.

Once again, this is a general rule, that the alternating sums of ๏Š๏Ž๐ถ are zero: ๏Š๏Šฆ๏Š๏Šง๏Š๏Šจ๏Š๏Šฑ๏Šง๏Š๏Š๏Šฑ๏Šง๏Š๏Š๏Š๐ถโˆ’๐ถ+๐ถโˆ’โ‹ฏ+(โˆ’1)(๐ถ)+(โˆ’1)(๐ถ)=0 or, more concisely, ๏Š๏Ž๏Šฒ๏Šฆ๏Ž๏Š๏Ž๏„š(โˆ’1)(๐ถ)=0.

An alternative way to represent this is that the sums of odd and even terms are equal. This is not surprising when ๐‘› is odd due to the reflective symmetry: ๏Š๏Ž๏Š๏Š๏Šฑ๏Ž๐ถ=๐ถ. However, as the previous example showed, this is also true for even ๐‘›.

Key Points

  1. The number of combinations of size ๐‘Ÿ taken from a set of size ๐‘› is given by ๏Š๏Ž๏Š๏Ž๐ถ=๐‘ƒ๐‘Ÿ!=๐‘›!๐‘Ÿ!(๐‘›โˆ’๐‘Ÿ)!.
  2. Combinations have the following key properties:
    • Reflective property: ๏Š๏Ž๏Š๏Š๏Šฑ๏Ž๐ถ=๐ถ,
    • Recursive property: ๏Š๏Šฑ๏Šง๏Ž๏Š๏Šฑ๏Šง๏Ž๏Šฑ๏Šง๏Š๏Ž๐ถ+๐ถ=๐ถ,
    • Summation: ๏Š๏Ž๏Šฒ๏Šฆ๏Š๏Ž๏Š๏„š๐ถ=2,
    • Alternating sum: ๏Š๏Ž๏Šฒ๏Šฆ๏Ž๏Š๏Ž๏„š(โˆ’1)(๐ถ)=0.
  3. Using the definition of ๏Š๏Ž๐ถ and its properties, we can simplify many expressions and solve equations involving combinations.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.