Question 1198404
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There are 6 choices for the first slot, then 6-1 = 5 choices for the second slot, and finally 5-1 = 4 choices for the third slot.


If order mattered, then we'd have 6*5*4 = 30*4 = 120 different permutations, and this would be the final answer.


But order doesn't matter. 
The group {A,B,C} is the same as something like {B,A,C}. 
The order doesn't matter when considering the group overall.


There are 3*2*1 = 6 ways to rearrange any given group of 3 items. 
This tells us that the previous figure of 120 is too large by a factor of 6.


Divide by 6 to correct for this over-counting: 120/6 = <font color=red>20</font>
Therefore, we have <font color=red>20</font> different groups of three letters chosen from a pool of six.


Alternatively you can use the nCr combination formula to get this answer. 
{{{nCr = (n!)/(r!*(n-r)!)}}}
Use n = 6 and r = 3. 


Another alternative is to use Pascal's Triangle. 
Look at the row that starts with 1,6,... and you're looking for the item in the fourth slot. 
This is because the 'r' value starts counting at r = 0 rather than r = 1.


Answer: <font color=red>20</font>
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