Permutations

Question

Evaluate the following permutations and combinations:

a.  P(5,3)

 

b.  C(5,3)

Answer 1

These two are the same except for one thing: in permutations (a), order matters. With combinations (b), the order of objects does not matter. For example, we choose 3 candies from a bag of 5 candies. Whether we get a caramel, a chocolate, then a mint, or a mint, a caramel, then a chocolate, it makes no difference - it is the same combination.

 

a. P(5,3) means the permutations of 5 objects chosen 3 at a time. The simplest way to think of this is that we draw three times & each time we have 1 fewer object to choose from. For example, I choose 1 out of 5, then 1 out the 4 that remain, and finally 1 out of the 3 that remain. Obviously, each choice changes the mix of objects available for the next draw. Let "n" represent the number of objects or choices, then "n-1" represents one less than that number, "n-2" is two less than the number of objects, etc.

 

n x (n-1) x (n-2) represents that we choose 3 objects (n, n-1, n-2) & we multiply the numbers of objects we can choose from.

 

So, 5x4x3=60. There are 60 permutations of 5 objects chosen 3 at a time, but there's a catch. This assumes that we cannot choose any object twice - that it is not returned to the pool of objects & there's no repetition of objects. Each object is unique among the group.

 

b. C(5,3) means combinations of 5 objects chosen 3 at a time, but now we aren't concerned about the order they are in. This starts out like permutations, but then we reduce the number by a factor that represents the many ordered arrangements of the same objects. For example, the previous 60 permutations include some groupings that have the same exact objects - but they are in different orders. Can we know how many "extra" combinations there are among these permutations? Yes! We use an interesting kind of multiplication called "factorial". Factorial simply means to multiply a series of descending natural numbers, and its sign is !.

 

If we choose 3 objects, then: 3! = 3 x 2 x 1 = 6 represents the factor by which we divide the number of permutations. In this case, we had 60 permutations. Divide that by 6 & we eliminate the 50 repetitive combinations among those permutations.

 

(5 x 4 x 3)/3! = 60/6 = 10

 

There are 10 combinations of 5 objects chosen 3 at a time. Again, we assume that each object is unique & there is no repetition. Otherwise, we have more material to cover!