A couple of years ago, my friends and I went to an indoor shooting range for a friend’s bachelor party. One of the firearms we chose was a .44 Magnum, a revolver made famous by Clint Eastwood’s “Dirty Harry” films. Holding the revolver in my hand, I thought, “Why is this weapon called a revolver? Doesn’t the cylinder rotate around its axis? Perhaps it would be more accurate to call this weapon a rotator.” I’ll admit, that’s probably not the first thing most people think about when they hold a revolver in their hand (besides, maybe the weapon is called a revolver because the bullets revolve about the cylinder’s axis).
In any case, the experience reminded me of another product that has a confusing (and arguably inaccurate) name, the combination lock. A combination lock is a type of lock in which you have to enter a specific sequence of numbers to open the lock. The name is confusing because, in mathematics, the correct term for a set of numbers that must be in a specific sequence is a permutation. While it probably wouldn’t be very productive to try to change the naming convention for the combination lock, it is important to understand the difference between combinations and permutations.
Deos Oedrr Mettar? Does Order Matter?
To know whether you’re dealing with a combination or a permutation, ask yourself one question – does order matter? If it does, you’re dealing with a permutation. If it doesn’t, you’re dealing with a combination. After determining whether order matters, you need to find out whether repetition is allowed in the selection process. For both combinations and permutations, the formulas are different depending on whether repetition is allowed.
In an earlier article, I mentioned I’m on a local policy advisory board and soon we’re going to elect three members of our 15-person board into leadership roles – Chair, Vice-chair, and 2nd Vice-chair. These are distinct roles, which are in decreasing order of responsibility. It’s fairly obvious that one outcome (e.g. Tess being elected Chair, Filbert being elected Vice-chair, and Laura being elected 2nd Vice-chair) is very different from another outcome (e.g. Laura being elected Chair, Tess being elected Vice-chair, and Filbert being elected 2nd Vice-chair).
Permutation – When Order Matters
Since order matters in this example, we know we’re dealing with a permutation, and since each person can only hold one position, we also know repetition is not allowed. Given this information, the number of ways we could select three members of our 15-person board for the ordered leadership positions is 2,730.1
Combination – When Order Doesn’t Matter
A similar example in which order does not matter is when we’re selecting a subset of members to be on one of the board’s sub-committees. In this case, a sub-committee composed of Tess, Filbert, and Laura is identical to a sub-committee composed of Laura, Tess, and Filbert. That is, the order of the members doesn’t matter. And since we can’t select a person twice for the same sub-committee, once again we know repetition is not allowed. Given this information, the number of ways we could select three members of our 15-person board for the unordered sub-committee positions is 455.2
|Does Order Matter?|
Reinforcing with Repetition
As you can see, it is important to determine whether order matters. In the first example, where it does, there are 2,730 permutations. In the second example, where it doesn’t, there are only 455 combinations. While we sometimes use the term combination loosely to describe sets of objects regardless of whether order matters, it’s important to remember that there is a difference between combinations and permutations and you can identify which one you’re dealing with by asking yourself whether the order of objects matters. By understanding the difference between the two concepts and being able to use the correct formulas, you’ll be able to calculate the right quantities and make informed decisions based on accurate information.
1 Permutation without repetition: n! / (n – r)! = 15! / (15 – 12)! = 15 * 14 * 13 = 2,730
2 Combinations without repetition: n! / r!(n – r)! = 15! / 3!(15 – 12)! = (15 * 14 * 13) / (3 * 2 * 1) = 455