On a recent post I went over factorials. Now let’s try a word problem that actually uses the factorial sign.

*Five students must sit in 5 chairs. How many different ways can they sit?*

(A) 30

(B) 60

(C) 4!

(D) 5!

(E) 6!

In this problem we have a set number of people. Our job is to find how many different ways to arrange them. This is what we call a permutations problem.

Two features of a permutations problem are as follows: the number of members in the group is fixed, and we are looking for the total possible ways to arrange those members.

To find the answer, we simply put the number in the group followed by a factorial sign: 5! = 5 x 4 x 3 x 2 x 1 = 120. In the question, you don’t have to write out the factorial and can directly answer (D).

Now what if we weren’t too concerned about who was sitting in a particular chair. Instead, we wanted to choose 5 students from a larger group—let’s say 8. How many different ways could we select 5 from a group of 8 to sit in the five chairs?

Notice that I am not using the word arrange, the way I did with a permutation problem. With a combinations problem the order of the people in the group (in this problem, the order in which they are sitting) does not matter. So if Steve and Mary are in the group of students to be seated, it doesn’t matter which chair they are sitting in.

The formula for a combinations problem is n!/(n – p)! p!, where n is the total number you are choosing from and p is the number you are choosing. In this problem, n = 8 (there are 8 members in the group) and p = 5 (we are choosing 5 students to sit in the five chairs.)

Plugging in these numbers we get:

8!/3!5! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1/(5 x 4 x 3 x 2 x 1) (3 x 2 x 1)

Notice that the 5 x 4 x 3 x 2 x 1 cancels out from both the numerator and the denominator. That leaves us with 8 x 7 x 6/3 x 2 x 1 = 56. That is, there are 56 different ways to organize the group.

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Since the order does not matter, why wont a simple 8 x 7 x 6 x 5 x 4 suffice?