Question 1209003
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This might be a bit strange, but I'll address the "order matters" portion first.


There are n = 8 letters and r = 6 slots to fill.
We have 8 choices for the first slot, then 7 for the next, and so on until reaching n-r+1 = 8-6+1 = 3 choices for the 6th slot.


Multiplying out these values gives: (8*7*6)*(5*4*3) = <font color=red>20160</font> different permutations where order matters.


Another way to reach this value is to use the nPr permutation formula.
nPr = (n!)/( (n-r)! )
which I'll let the student handle the scratch work if s/he chooses this route.
The exclamation mark indicates a factorial.


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Now onto the "order doesn't matter" portion.
For any group of 6 items, we have 6! = 6*5*4*3*2*1 = 720 ways to rearrange said group.
This means that the previous value we found (20160) is too large by a factor of 720 if order doesn't matter.
We'll divide the permutation value over r! = 6! = 720 to find the combination value.


A very useful formula is
nCr = (nPr)/(r!)
which is a way to connect permutations to combinations. 

There are 20160/720 = <font color=red>28</font> combinations.


This value can be found by looking at Pascal's Triangle. 
Since n = 8, we start with the row "1,8,..." 
r = 0 is the starting index and refers to the left-most value in this row. Increment r by 1 each time you move to the right. Once reaching r = 6, you should arrive at 28 in Pascal's Triangle.


Or if you wanted you can use this nCr formula
nCr = (n!)/( r!*(n-r)! )
I'll let the student handle the scratch work if s/he chooses this route.
Notice the presence of n! up top and (n-r)! down below; both of which are found in the nPr formula. 
So that can help explain why nCr = (nPr)/(r!) is true.


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Answers:


How many ways can this be done, if...
Order doesn't matter?  <font color=red>28</font>
Order does matter?  <font color=red>20160</font>
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