Question 875415
<pre>
The chairs are ordered 1 through 9.

Suppose the students are Adam, Betty, Calvin, Diane, Edward, and Frances.

Order matters, so it is a permutation.  

We can assign Adam a seat in any of the 9 chairs.  
For each of those 9 ways we can seat Adam,
there remain any of 8 chairs to assign to Betty.
So there are 9*8 or 72 ways to seat Adam and Betty.

For each of those 9*8 or 72 ways we can seat Adam and Betty,
there remain any of 7 chairs to assign to Calvin.
So there are 9*8*7 or 504 ways to seat Adam, Betty, and Calvin.

For each of those 9*8*7 or 504 ways we can seat Adam, Betty and Calvin,
there remain any of 6 chairs to assign to Diane.
So there are 9*8*7*6 or 3024 ways to seat Adam, Betty, Calvin and Diane.

For each of those 9*8*7*6 or 3024 ways we can seat Adam, Betty, Calvin,
and Diane,
there remain any of 5 chairs to assign to Edward.
So there are 9*8*7*6*5 or 15120 ways to seat Adam, Betty, Calvin, Diane
and Edward.

For each of those 9*8*7*6*5 or 15120 ways we can seat Adam, Betty, Calvin,
Diane and Edward,
there remain any of 4 chairs to assign to Frances.
So there are 9*8*7*6*5*4 or 60480 ways to seat Adam, Betty, Calvin, Diane,
Edward and Frances.

Answer: 9*8*7*6*5*4 = 60480 

Different people label that different ways.  Here are the common ways: 

"The number of permutations of 9 things taken 6 at a time" =

"9P6" = "9 position 6" = P(9,6) = "{{{(matrix(2,1,9,6))}}}".

You can get it on a calculator:
 
On a TI-83 or 84

Press ON
Press CLEAR
Press 9
Press MATH
Press the left arrow once
Press 2
Press 6
You'll see this:   9 nPr 6
Press ENTER
You'll see this:   9 nPr 6     
                            60480

Edwin</pre>