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Very confused on these particular clue what it's talking about...

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@ganeshie8 I figured out C(x), but I don't get P(x)
For 16th question, i think P(X) = 4C2

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Other answers:

the two element group...say there are 4 elements,ABCD,there will be 16 2 element grp,AA,AB,AC,....DD....but if repetition like AA are not allowed,then there will be 12...and if AB and BA are considered same ,then there will be 6....
I am getting B for the first one.
Because I did xP2 = P(x) and (n - 1)! = C(x)
Or would it be xC2 = P(x)?
oh i see... but i guess forming 2 element groups involves just choosing elements, order doesnt matter...
Oh ok...then I get D for 16 and B for 17. I ddin't realize that, I was still thinking about the other part of the problem.
if order doe not matter,then AB and BA are same,then use xC2
Ok, then it would be: 16. D 17. B ?
yeah i guess that it.... .
Alright. Thanks :)
np... :)

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