Teaching Permutations and Combinations with Sports
In this edition of Teaching Math with Sports, we look at applications of permutations and combinations in sports. Permutations and combinations are two of the main topics under the (more complicated than it sounds) umbrella of counting.
Permutations and combinations are used to answer questions that start with “how many ….?”. The typical applications seen in high school and college undergrad statistics classes deal with playing cards or rolling dice. For me, these applications were always particularly uncompelling. That is, it was hard to care about permutations and combinations because I didn’t care about the examples.
In this article, we’ll try to fix that problem by showing how combinations and permutations can be applied in sports.
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Background Theory: Permutations and Combinations
As I mentioned above, studying permutations and combinations is really about studying how to answer questions that begin with “how many”. An example that showed up on an AP statistics exam a few years ago asked “how many different 5 card poker hands are there that are a straight and contain a 5?”
Problems in this form are commonly called counting problems. I like to think of them more accurately as enumeration problems. Counting is something you do in kindergarten – enumeration is a more mature subject. No matter what you call the subject, though, almost every problem in this domain can be answered using permutations and combinations.
Both combinations and permutations deal with selecting groups of a fixed size from a set of candidate choices. In both cases, repetitions are not allowed in picking the group from the set of candidates. How do they differ? That is, a candidate can only be selected once. Sometimes, permutations and combinations are referred to as “selecting without replacement”.
Suppose we have n candidates and a group of size k to select. The number of combinations in this setting is the number of ways to choose a group of size k from n candidate choices. The number of permutations is the number of ways to do this where order matters.
For example, suppose we had to pick 4 numbers from the set \{1,\dots, 10\} . The choices 1, 2, 3, 4 and 4, 1, 2, 3 are considered the same from a combinations perspective but differente from a permutations perspective.
A pretty good example comes from picking committees in a governing body. If we want to pick 3 members of the 100 person US senate to serve on a committee, then this is a “combination” problem. If we want to pick 3 members of the 100 person US senate to serve as the chair, vice chair, and treasurer on a committee, this is a permutation problem. The difference is that, though in both cases we’re picking 3 from 100, in the first example the 3 members are indistinguishable while in the second case the order matters.
Numbers of combinations are computed using the “choose” function. For example, the number of 3 person groups chosen from amongst 10 candidates is given by the function “10 choose 3”, denoted 10C3.
On the other hand, permutations are computed using “permute”. In the same example the number of permutations of 3 people from 10 candidates is given by “10 permute 3”, denoted 10P3.
The formulas for both nCk and nPk are given in terms of factorials. “n factorial” is denoted n! and has the formula n!=n \cdot (n-1) \cdot (n-2)\dots\cdot 2\cdot 1 .
In this notation we have nCk = \frac{n!}{k!(n-k)!} and nPk = \frac{n!}{(n-k)!} .
Example 1: Number of Possible Basketball Lineups
In the NBA, teams are allowed to roster 15 different players. However, only 5 players are allowed on the court at a time. This is a perfect opportunity to apply combinations and permutations.
First, the total number of 5 person teams you can form out of 15 players is a combination problem. The parameters are n= 15 and k=5 . The number of 5 person teams is given by 15C5 = \frac{15!}{10!5!}=3003 . That means that on a given night, there are 3003 different starting rosters that an NBA team could come out onto the court with.
However, when we consider positions, the number gets even bigger. What if we want to count the number of 5 person teams out of 15 possible players where teams with the same players at different positions count as different teams? This is a permutation problem. The answer here is given by 15P5 = \frac{15!}{10!}=360360.
Challenge Question 1: Suppose there are 8 players who can play guard and 7 who can play as a big (PF/C). If we start 3 guards and 2 bigs, how many starting lineups are there? If we start 2 guards and 3 bigs how many lineups are there?
Super Challenge Question 2: Suppose our team has 6 players who can only play guard, 6 guys who can only play big, and 3 guys who can play both. How would you answer the questions posed in challenge question 2?
Example 2: NFL Draft and Probabilities
Permutations and combinations can also be used to calculate probabilities when everything is “equally likely”. Let’s use the NFL draft as an example.
Suppose that for picks 11-20 in an NFL draft, there are 18 offensive players and 16 defensive players that are all equally likely to be taken in this range. We want to estimate the probability that the same number of offensive players are taken as defensive players in this range. That is, what is the probability that of the 18 offensive players and 16 defensive players, 5 of each were drafted.
Believe it or not this is a counting problem.
Counting problems turn into probabilities as follows. We want to determine the probability that something happens in a particular way. In our case, 10 players are drafted with 5 offensive and 5 defensive players. First we count the number of 10 person groups that satisfy this constraint. Then we count the total number of 10 person groups that can be chosen from the total 34 candidates. The first number divided by the second number is our probability.
Notice that we didn’t ask about order, just about selecting 10 players from the given candidates. This is a combination problem. In order to pick 10 players with 5 from offense and 5 from defense, we can pick 5 of the 18 offensive players and 5 of the 16 defensive players. Then, the product of these two numbers is the number we’re looking for. Specifically, the number of ways to pick 5 offensive and defensive players is (18C5)\cdot(16C5)=8568\cdot 4368=37,425,024.
Now, the total number of ways that a 10 person group can be chosen from the 34 candidates regardless of position is 34C10=131,138,140 . Then, the probability of equal numbers of offensive and defensive players being drafted is \frac{37,425,024}{131,138,140}=28.5%.
Challenge Question 3: How would you compute the probability that more offensive players were taken than defensive players?
Example 3: Rediscovering Factorials with Baseball Lineups
Permutations and combinations are related through whether or not ordering matters in the group. There are always more permutations than combinations because for every group that can be chosen, there are multiple ways in which it can be ordered. But how many more permutations are there than combinations?
Let us work an example in baseball. Baseball rosters have 26 players, of which 9 are in the batting order. However, perhaps only 15 of those players will be hitters. Therefore, the number of 9-person lineups without order is 15C9 = 5005 . However, when order matters (as it naturally does in baseball lineups!), the number of 9-person lineups is 15P9 = 1816214400.
This number is astronomically larger! If we divide 15P9 by 15C9 , you get 362880 . This number may seem inconspicuous, but 362880 = 9\cdot 8\cdot \dots \cdot 1 = 9!. That means that for every combination of size k there are exactly k! permutations, or possible ways to order the k players.
You can verify this fact in general by dividing the formulas for permutations and combinations and seeing the factorial pop out.
Challenge Question Answers:
- For the first two questions, you can only choose the guards from the pool of guards and the bigs from the pool of bigs. Then, these combinations are multiplied together to yield the number of teams.
- In the other case, first pick the guards from the pool of potential guards. This includes the guards and the players that can play both ways. Then, we need to condition on the number of both-way players selected as guards to determine how many choices we have for bigs.
- Add together the probability of 6, 7, 8. 9, or 10 offensive players, each computed in the same way as described in the example.
