Before working through this lesson on "combinations" the user should look back through the lesson and practice page on "permutations." 
Many of the same ideas and symbols will be used in this lesson. 
The major difference between a permutation and a combination is that in a combination the order is not important, whereas in a permutation the order of the arrangement is important.
For example, if you were asked how many permutations of the letters a,b and c there are you would come up with 6 (abc, acb,bac,bca,cab, and cba are all different permutations)
but if you asked how many combinations of those
same letters there are, you would come up
with just 1 combination( abc,acb,bac,bca,cab, and cba are all considered to be the same combination because the order they appear is not important)
 

Let's learn the rules!

Before we go any further we need another important definition.
In this lesson we will see numbers, followed by what looks like an exclamation point.
They look like this..... 4!
No, this does not mean to yell the number 4.
In mathematics this is called a factorial sign, and so this would be said "4 factorial."
A factorial works like this:
4! = 4X3X2X1 = 24
6! = 6X5X4X3X2X1 = 720
or in other words
n! = (n)(n-1)(n-2)(n-3)....(1)
 

Ok...now for the rules!!

_nC_n

Combinations of "n" things taken "n" at a time.

Is always  1

_nC_r

Combinations of "n" things taken "r" at a time
(r is always less than n)

 In this problem:
n=5
r=3
so (n-r) = (5-3)=2
Now just follow the rule:

The answer is 10 combinations.

    This is a combination problem because who attends the meeting is not important.
In this problem:
"n" = 12
"r" = 2
Plug these into the formula:










 

After we cancel all the common factors the
number of combinations is 66!
66 different pairs of kids could be
chosen to attend the meeting.

The trick in all of these problems is to simplify it down to the basic formula.  In this case, all of the words can really be replaced with this simple formula:
₇C₃ 
(there are 7 letters in "Algebra")
Plug in the values:
n = 7
r = 3

After all the multiplying and canceling, the result is:
35 - 3 letter combinations can be formed.