There are total of $16$ people. How many ways can you make two committees of $8$ people in each committee? You have $16$ people in a class. How many ways can you make two committees of $8$ people in each committee?

How do I do this problem?
How many ways are there to choose the first person on your committee? 16. How about the second person? 15, because there is one person who is not available. So we have $16*15*14*13*12*11*10*9$ ways. But we've overcounted because we've chosen the same committees of people in several different orders. How many times did we count each committee? This is $8!=8*7*6*5*4*3*2*1$ because there are this many ways to order $8$ people. (8 ways to choose the first, 7 ways to choose the second...) So we divide $16*15*14*13*12*11*10*9$ by $8!=8*7*6*5*4*3*2*1$ to account for the overcounting. Assuming that we don't care to which committee a certain group of $8$ people is assigned, we must also divide by 2 because it's the same to choose a group of people or its complement. 

The final answer is $\frac{16*15*14*13*12*11*10*9}{8*7*6*5*4*3*2*1}= \frac{16!}{2*8!8!}=\frac{1}{2}\binom{16}{8}$.
An alternative way of thinking about this problem is that you line 16 people up in a line where each position decides what committee they will be placed on. Then we must divide out the redundant combinations. so $ \frac{16!}{8!*8!} $